Evaluate $\displaystyle \lim_{x \to 1}\dfrac{\left( \log \left( 1+x \right)-\log 2 \right)\left( 3\cdot {{4}^{x-1}}-3x \right)}{\left\{ {{\left( 7+x \right)}^{\dfrac{1}{3}}}-{{\left( 1+3x \right)}^{\dfrac{1}{2}}} \right\}\sin \pi x}$ .
Answer
Verified521.7k+ views
Hint: We have to substitute $x=1+h$ . When $x \to 1$ , we can write from this equation that $h\to 0$ . Then, we have to simplify the expression using logarithmic rules. We have to convert the expression in such a way that we can use the formulas $\displaystyle \lim_{x \to 0}\dfrac{\log \left( 1+x \right)}{x}=1$ and $\displaystyle \lim_{x \to 0}\dfrac{\sin x}{x}=1$ . We have to apply the L-Hospitals rule to simplify further.
Complete step-by-step solution:
We have to evaluate $\displaystyle \lim_{x \to 1}\dfrac{\left( \log \left( 1+x \right)-\log 2 \right)\left( 3\cdot {{4}^{x-1}}-3x \right)}{\left\{ {{\left( 7+x \right)}^{\dfrac{1}{3}}}-{{\left( 1+3x \right)}^{\dfrac{1}{2}}} \right\}\sin \pi x}$ .
Let us substitute $x=1+h$ . When $x \to 1$ , we can write from this equation that $h\to 0$ . Let us substitute the values in the given expression.
$\begin{align}
& \Rightarrow \displaystyle \lim_{h\to 0}\dfrac{\left( \log \left( 1+1+h \right)-\log 2 \right)\left( 3\cdot {{4}^{1+h-1}}-3\left( 1+h \right) \right)}{\left\{ {{\left( 7+1+h \right)}^{\dfrac{1}{3}}}-{{\left( 1+3\left( 1+h \right) \right)}^{\dfrac{1}{2}}} \right\}\sin \pi \left( 1+h \right)} \\
& \Rightarrow \displaystyle \lim_{h\to 0}\dfrac{\left( \log \left( 2+h \right)-\log 2 \right)\left( 3\cdot {{4}^{h}}-3-3h \right)}{\left\{ {{\left( 8+h \right)}^{\dfrac{1}{3}}}-{{\left( 1+3\left( 1+h \right) \right)}^{\dfrac{1}{2}}} \right\}\sin \pi \left( 1+h \right)} \\
\end{align}$
We know that $\log a-\log b=\log \left( \dfrac{a}{b} \right)$ . Therefore, we can write the above equation as
$\Rightarrow \displaystyle \lim_{h\to 0}\dfrac{\log \left( \dfrac{2+h}{2} \right)\left( 3\cdot {{4}^{h}}-3-3h \right)}{\left\{ {{\left( 8+h \right)}^{\dfrac{1}{3}}}-{{\left( 1+3\left( 1+h \right) \right)}^{\dfrac{1}{2}}} \right\}\sin \pi \left( 1+h \right)}$
We can write $\left( \dfrac{2+h}{2} \right)$ as $1+\dfrac{h}{2}$ .
$\begin{align}
& \Rightarrow \displaystyle \lim_{h\to 0}\dfrac{\log \left( 1+\dfrac{h}{2} \right)\left( 3\cdot {{4}^{h}}-3-3h \right)}{\left\{ {{\left( 8+h \right)}^{\dfrac{1}{3}}}-{{\left( 1+3\left( 1+h \right) \right)}^{\dfrac{1}{2}}} \right\}\sin \pi \left( 1+h \right)} \\
& \Rightarrow \displaystyle \lim_{h\to 0}\dfrac{\log \left( 1+\dfrac{h}{2} \right)\left( 3\cdot {{4}^{h}}-3-3h \right)}{\left\{ {{\left( 8+h \right)}^{\dfrac{1}{3}}}-{{\left( 1+3+3h \right)}^{\dfrac{1}{2}}} \right\}\sin \pi \left( 1+h \right)} \\
& \Rightarrow \displaystyle \lim_{h\to 0}\dfrac{\log \left( 1+\dfrac{h}{2} \right)\left( 3\cdot {{4}^{h}}-3-3h \right)}{\left\{ {{\left( 8+h \right)}^{\dfrac{1}{3}}}-{{\left( 4+3h \right)}^{\dfrac{1}{2}}} \right\}\sin \pi \left( 1+h \right)} \\
\end{align}$
Let us apply distributive property on $\sin \pi \left( 1+h \right)$ .
$\Rightarrow \displaystyle \lim_{h\to 0}\dfrac{\log \left( 1+\dfrac{h}{2} \right)\left( 3\cdot {{4}^{h}}-3-3h \right)}{\left\{ {{\left( 8+h \right)}^{\dfrac{1}{3}}}-{{\left( 4+3h \right)}^{\dfrac{1}{2}}} \right\}\sin \left( \pi +\pi h \right)}$
We know that $\sin \left( a+b \right)=\sin a\cos b+\cos a\sin b$ . Therefore, the above equation becomes
$\Rightarrow \displaystyle \lim_{h\to 0}\dfrac{\log \left( 1+\dfrac{h}{2} \right)\left( 3\cdot {{4}^{h}}-3-3h \right)}{\left\{ {{\left( 8+h \right)}^{\dfrac{1}{3}}}-{{\left( 4+3h \right)}^{\dfrac{1}{2}}} \right\}\left( \sin \pi \cos \pi h+\cos \pi \sin \pi h \right)}$
We know that $\sin \pi =0$ and $\cos \pi =-1$ . We can write the above equation as
$\begin{align}
& \Rightarrow \displaystyle \lim_{h\to 0}\dfrac{\log \left( 1+\dfrac{h}{2} \right)\left( 3\cdot {{4}^{h}}-3-3h \right)}{\left\{ {{\left( 8+h \right)}^{\dfrac{1}{3}}}-{{\left( 4+3h \right)}^{\dfrac{1}{2}}} \right\}\left( 0\times \cos \pi h+-1\times \sin \pi h \right)} \\
& \Rightarrow \displaystyle \lim_{h\to 0}\dfrac{\log \left( 1+\dfrac{h}{2} \right)\left( 3\cdot {{4}^{h}}-3-3h \right)}{-\left\{ {{\left( 8+h \right)}^{\dfrac{1}{3}}}-{{\left( 4+3h \right)}^{\dfrac{1}{2}}} \right\}\sin \pi h} \\
\end{align}$
Let us multiply and divide the denominator by $\pi h$ .
$\Rightarrow \displaystyle \lim_{h\to 0}\dfrac{\log \left( 1+\dfrac{h}{2} \right)\left( 3\cdot {{4}^{h}}-3-3h \right)}{-\left\{ {{\left( 8+h \right)}^{\dfrac{1}{3}}}-{{\left( 4+3h \right)}^{\dfrac{1}{2}}} \right\}\dfrac{\sin \pi h}{\pi h}\times \pi h}$
We have to multiply and divide the denominator by 2.
\[\Rightarrow \displaystyle \lim_{h\to 0}\dfrac{\log \left( 1+\dfrac{h}{2} \right)\left( 3\cdot {{4}^{h}}-3-3h \right)}{-\left\{ {{\left( 8+h \right)}^{\dfrac{1}{3}}}-{{\left( 4+3h \right)}^{\dfrac{1}{2}}} \right\}\dfrac{\sin \pi h}{\pi h}\times \dfrac{\pi h}{2}\times 2}\]
We know that $\displaystyle \lim_{x \to a}\dfrac{f\left( x \right)}{g\left( x \right)}=\dfrac{\displaystyle \lim_{x \to a}f\left( x \right)}{\displaystyle \lim_{x \to a}g\left( x \right)}$ and $\displaystyle \lim_{x \to a}f\left( x \right)g\left( x \right)=\displaystyle \lim_{x \to a}f\left( x \right)\times \displaystyle \lim_{x \to a}g\left( x \right)$ .
$\Rightarrow \dfrac{\displaystyle \lim_{h\to 0}\dfrac{\log \left( 1+\dfrac{h}{2} \right)}{\dfrac{h}{2}}}{\displaystyle \lim_{h\to 0}\dfrac{\sin \pi h}{\pi h}}\displaystyle \lim_{h\to 0}\dfrac{\left( 3\cdot {{4}^{h}}-3-3h \right)}{-\left\{ {{\left( 8+h \right)}^{\dfrac{1}{3}}}-{{\left( 4+3h \right)}^{\dfrac{1}{2}}} \right\}2\pi }$
We know that $\displaystyle \lim_{x \to 0}\dfrac{\log \left( 1+x \right)}{x}=1$ and $\displaystyle \lim_{x \to 0}\dfrac{\sin x}{x}=1$ .
$\begin{align}
& \Rightarrow 1\times \displaystyle \lim_{h\to 0}\dfrac{\left( 3\cdot {{4}^{h}}-3-3h \right)}{-\left\{ {{\left( 8+h \right)}^{\dfrac{1}{3}}}-{{\left( 4+3h \right)}^{\dfrac{1}{2}}} \right\}2\pi } \\
& \Rightarrow \displaystyle \lim_{h\to 0}\dfrac{\left( 3\cdot {{4}^{h}}-3-3h \right)}{-\left\{ {{\left( 8+h \right)}^{\dfrac{1}{3}}}-{{\left( 4+3h \right)}^{\dfrac{1}{2}}} \right\}2\pi } \\
\end{align}$
We have to take the common factor 3 outside from the numerator.
$\Rightarrow \displaystyle \lim_{h\to 0}\dfrac{3\left( {{4}^{h}}-1-h \right)}{-\left\{ {{\left( 8+h \right)}^{\dfrac{1}{3}}}-{{\left( 4+3h \right)}^{\dfrac{1}{2}}} \right\}2\pi }$
We can see that when we apply the limit, we will get the result in $\dfrac{0}{0}$ . Therefore, we have to apply L'Hospital's Rule. According to this rule, we will find the derivative of the numerator and the denominator.
$\Rightarrow \displaystyle \lim_{h\to 0}\dfrac{\dfrac{d}{dh}\left\{ 3\left( {{4}^{h}}-1-h \right) \right\}}{\dfrac{d}{dh}\left[ -2\pi \left\{ {{\left( 8+h \right)}^{\dfrac{1}{3}}}-{{\left( 4+3h \right)}^{\dfrac{1}{2}}} \right\} \right]}$
We know that $\dfrac{d}{dx}ax=a\dfrac{d}{dx}x$ .
$\begin{align}
& \Rightarrow \displaystyle \lim_{h\to 0}\dfrac{3\dfrac{d}{dh}\left( {{4}^{h}}-1-h \right)}{-2\pi \dfrac{d}{dh}\left\{ {{\left( 8+h \right)}^{\dfrac{1}{3}}}-{{\left( 4+3h \right)}^{\dfrac{1}{2}}} \right\}} \\
& \Rightarrow \displaystyle \lim_{h\to 0}\dfrac{3\left[ \dfrac{d}{dh}{{4}^{h}}-\dfrac{d}{dh}1-\dfrac{d}{dh}h \right]}{-2\pi \left[ \dfrac{d}{dh}{{\left( 8+h \right)}^{\dfrac{1}{3}}}-\dfrac{d}{dh}{{\left( 4+3h \right)}^{\dfrac{1}{2}}} \right]}...\left( i \right) \\
\end{align}$
Let us consider $y={{4}^{h}}...\left( iii \right)$ . We have to take logarithms on both sides.
$\Rightarrow \log y=\log {{4}^{h}}$
We know that $\log {{x}^{a}}=a\log x$ .
$\Rightarrow \log y=h\log 4$
Let us differentiate both the sides with respect to h. We know that $\dfrac{d}{dx}{{\log }_{e}}x=\dfrac{1}{x}$ and $\dfrac{d}{dx}ax=a\dfrac{d}{dx}x$ . Therefore, the above equation becomes
$\begin{align}
& \Rightarrow \dfrac{1}{y}\dfrac{dy}{dh}={{\log }_{e}}4 \\
& \Rightarrow \dfrac{dy}{dh}=y{{\log }_{e}}4 \\
\end{align}$
We have to substitute (iii) in the above equation.
$\Rightarrow \dfrac{dy}{dh}={{4}^{h}}{{\log }_{e}}4$
Let us substitute the above value in the equation (i).
$\Rightarrow \displaystyle \lim_{h\to 0}\dfrac{3\left[ {{4}^{h}}{{\log }_{e}}4-\dfrac{d}{dh}1-\dfrac{d}{dh}h \right]}{-2\pi \left[ \dfrac{d}{dh}{{\left( 8+h \right)}^{\dfrac{1}{3}}}-\dfrac{d}{dh}{{\left( 4+3h \right)}^{\dfrac{1}{2}}} \right]}$
We know that $\dfrac{d}{dx}a=0$ and $\dfrac{d}{dx}{{x}^{n}}=n{{x}^{n-1}}$ . Therefore, the above equation becomes
$\begin{align}
& \Rightarrow \displaystyle \lim_{h\to 0}\dfrac{3\left[ {{4}^{h}}{{\log }_{e}}4-0-1 \right]}{-2\pi \left[ \dfrac{1}{3}{{\left( 8+h \right)}^{\dfrac{1}{3}-1}}-\dfrac{1}{2}{{\left( 4+3h \right)}^{\dfrac{1}{2}-1}}\times 3 \right]} \\
& \Rightarrow \displaystyle \lim_{h\to 0}\dfrac{3\left[ {{4}^{h}}{{\log }_{e}}4-1 \right]}{-2\pi \left[ \dfrac{1}{3}{{\left( 8+h \right)}^{-\dfrac{2}{3}}}-\dfrac{3}{2}{{\left( 4+3h \right)}^{-\dfrac{1}{2}}} \right]} \\
\end{align}$
Now, let us apply the limit.
$\begin{align}
& \Rightarrow \dfrac{3\left[ {{4}^{0}}{{\log }_{e}}4-1 \right]}{-2\pi \left[ \dfrac{1}{3}{{\left( 8+0 \right)}^{-\dfrac{2}{3}}}-\dfrac{3}{2}{{\left( 4+0 \right)}^{-\dfrac{1}{2}}} \right]} \\
& \Rightarrow \dfrac{3\left[ \log {{4}_{e}}-1 \right]}{-2\pi \left[ \dfrac{1}{3}\cdot {{8}^{-\dfrac{2}{3}}}-\dfrac{3}{2}\cdot {{4}^{-\dfrac{1}{2}}} \right]}...\left( ii \right) \\
\end{align}$
Let us evaluate ${{8}^{-\dfrac{2}{3}}}$ .
$\Rightarrow {{8}^{-\dfrac{2}{3}}}=\dfrac{1}{{{8}^{\dfrac{2}{3}}}}$
We know that ${{\left( {{a}^{m}} \right)}^{n}}={{a}^{mn}}$ .
$\Rightarrow {{8}^{-\dfrac{2}{3}}}=\dfrac{1}{{{\left( {{8}^{2}} \right)}^{\dfrac{1}{3}}}}$
$\Rightarrow {{8}^{-\dfrac{2}{3}}}=\dfrac{1}{\sqrt[3]{{{8}^{2}}}}$
$\Rightarrow {{8}^{-\dfrac{2}{3}}}=\dfrac{1}{\sqrt[3]{64}}$
$\Rightarrow {{8}^{-\dfrac{2}{3}}}=\dfrac{1}{4}...\left( iii \right)$
Now, let us find ${{4}^{-\dfrac{1}{2}}}$ .
$\Rightarrow {{4}^{-\dfrac{1}{2}}}=\dfrac{1}{\sqrt{4}}=\dfrac{1}{2}...\left( iv \right)$
We have to substitute (iii) and (iv) in equation (ii).
$\begin{align}
& \Rightarrow \dfrac{3\left[ {{\log }_{e}}4-1 \right]}{-2\pi \left[ \dfrac{1}{3}\cdot \dfrac{1}{4}-\dfrac{3}{2}\cdot \dfrac{1}{2} \right]} \\
& =\dfrac{3\left[ {{\log }_{e}}4-1 \right]}{-2\pi \left[ \dfrac{1}{12}-\dfrac{3}{4} \right]} \\
& =\dfrac{3\left[ {{\log }_{e}}4-1 \right]}{-2\pi \times \dfrac{-8}{12}} \\
& =\dfrac{3\left[ {{\log }_{e}}4-1 \right]}{-2\pi \times \dfrac{-2}{3}} \\
\end{align}$
Let us simplify the above expression.
$\begin{align}
& \Rightarrow \dfrac{3\left[ {{\log }_{e}}4-1 \right]\times 3}{4\pi } \\
& =\dfrac{9\left[ {{\log }_{e}}4-1 \right]}{4\pi } \\
\end{align}$
We know that ${{\log }_{e}}e=1$ . Therefore, we can write the above equation as
$\Rightarrow \dfrac{9\left[ {{\log }_{e}}4-{{\log }_{e}}e \right]}{4\pi }$
We know that $\log a-\log b=\log \left( \dfrac{a}{b} \right)$ . Therefore, we can write the above equation as
\[\begin{align}
& \Rightarrow \dfrac{9{{\log }_{e}}\left( \dfrac{4}{e} \right)}{4\pi } \\
& =\dfrac{9}{4\pi }{{\log }_{e}}\left( \dfrac{4}{e} \right) \\
\end{align}\]
Hence, the value of $\displaystyle \lim_{x \to 1}\dfrac{\left( \log \left( 1+x \right)-\log 2 \right)\left( 3\cdot {{4}^{x-1}}-3x \right)}{\left\{ {{\left( 7+x \right)}^{\dfrac{1}{3}}}-{{\left( 1+3x \right)}^{\dfrac{1}{2}}} \right\}\sin \pi x}$ is \[\dfrac{9}{4\pi }{{\log }_{e}}\left( \dfrac{4}{e} \right)\] .
Note: Students must be thorough with logarithmic rules and properties. They must deeply learn rules associated with limits and how to evaluate limits using L-Hospitals rule. Students must know to find the derivatives.
Complete step-by-step solution:
We have to evaluate $\displaystyle \lim_{x \to 1}\dfrac{\left( \log \left( 1+x \right)-\log 2 \right)\left( 3\cdot {{4}^{x-1}}-3x \right)}{\left\{ {{\left( 7+x \right)}^{\dfrac{1}{3}}}-{{\left( 1+3x \right)}^{\dfrac{1}{2}}} \right\}\sin \pi x}$ .
Let us substitute $x=1+h$ . When $x \to 1$ , we can write from this equation that $h\to 0$ . Let us substitute the values in the given expression.
$\begin{align}
& \Rightarrow \displaystyle \lim_{h\to 0}\dfrac{\left( \log \left( 1+1+h \right)-\log 2 \right)\left( 3\cdot {{4}^{1+h-1}}-3\left( 1+h \right) \right)}{\left\{ {{\left( 7+1+h \right)}^{\dfrac{1}{3}}}-{{\left( 1+3\left( 1+h \right) \right)}^{\dfrac{1}{2}}} \right\}\sin \pi \left( 1+h \right)} \\
& \Rightarrow \displaystyle \lim_{h\to 0}\dfrac{\left( \log \left( 2+h \right)-\log 2 \right)\left( 3\cdot {{4}^{h}}-3-3h \right)}{\left\{ {{\left( 8+h \right)}^{\dfrac{1}{3}}}-{{\left( 1+3\left( 1+h \right) \right)}^{\dfrac{1}{2}}} \right\}\sin \pi \left( 1+h \right)} \\
\end{align}$
We know that $\log a-\log b=\log \left( \dfrac{a}{b} \right)$ . Therefore, we can write the above equation as
$\Rightarrow \displaystyle \lim_{h\to 0}\dfrac{\log \left( \dfrac{2+h}{2} \right)\left( 3\cdot {{4}^{h}}-3-3h \right)}{\left\{ {{\left( 8+h \right)}^{\dfrac{1}{3}}}-{{\left( 1+3\left( 1+h \right) \right)}^{\dfrac{1}{2}}} \right\}\sin \pi \left( 1+h \right)}$
We can write $\left( \dfrac{2+h}{2} \right)$ as $1+\dfrac{h}{2}$ .
$\begin{align}
& \Rightarrow \displaystyle \lim_{h\to 0}\dfrac{\log \left( 1+\dfrac{h}{2} \right)\left( 3\cdot {{4}^{h}}-3-3h \right)}{\left\{ {{\left( 8+h \right)}^{\dfrac{1}{3}}}-{{\left( 1+3\left( 1+h \right) \right)}^{\dfrac{1}{2}}} \right\}\sin \pi \left( 1+h \right)} \\
& \Rightarrow \displaystyle \lim_{h\to 0}\dfrac{\log \left( 1+\dfrac{h}{2} \right)\left( 3\cdot {{4}^{h}}-3-3h \right)}{\left\{ {{\left( 8+h \right)}^{\dfrac{1}{3}}}-{{\left( 1+3+3h \right)}^{\dfrac{1}{2}}} \right\}\sin \pi \left( 1+h \right)} \\
& \Rightarrow \displaystyle \lim_{h\to 0}\dfrac{\log \left( 1+\dfrac{h}{2} \right)\left( 3\cdot {{4}^{h}}-3-3h \right)}{\left\{ {{\left( 8+h \right)}^{\dfrac{1}{3}}}-{{\left( 4+3h \right)}^{\dfrac{1}{2}}} \right\}\sin \pi \left( 1+h \right)} \\
\end{align}$
Let us apply distributive property on $\sin \pi \left( 1+h \right)$ .
$\Rightarrow \displaystyle \lim_{h\to 0}\dfrac{\log \left( 1+\dfrac{h}{2} \right)\left( 3\cdot {{4}^{h}}-3-3h \right)}{\left\{ {{\left( 8+h \right)}^{\dfrac{1}{3}}}-{{\left( 4+3h \right)}^{\dfrac{1}{2}}} \right\}\sin \left( \pi +\pi h \right)}$
We know that $\sin \left( a+b \right)=\sin a\cos b+\cos a\sin b$ . Therefore, the above equation becomes
$\Rightarrow \displaystyle \lim_{h\to 0}\dfrac{\log \left( 1+\dfrac{h}{2} \right)\left( 3\cdot {{4}^{h}}-3-3h \right)}{\left\{ {{\left( 8+h \right)}^{\dfrac{1}{3}}}-{{\left( 4+3h \right)}^{\dfrac{1}{2}}} \right\}\left( \sin \pi \cos \pi h+\cos \pi \sin \pi h \right)}$
We know that $\sin \pi =0$ and $\cos \pi =-1$ . We can write the above equation as
$\begin{align}
& \Rightarrow \displaystyle \lim_{h\to 0}\dfrac{\log \left( 1+\dfrac{h}{2} \right)\left( 3\cdot {{4}^{h}}-3-3h \right)}{\left\{ {{\left( 8+h \right)}^{\dfrac{1}{3}}}-{{\left( 4+3h \right)}^{\dfrac{1}{2}}} \right\}\left( 0\times \cos \pi h+-1\times \sin \pi h \right)} \\
& \Rightarrow \displaystyle \lim_{h\to 0}\dfrac{\log \left( 1+\dfrac{h}{2} \right)\left( 3\cdot {{4}^{h}}-3-3h \right)}{-\left\{ {{\left( 8+h \right)}^{\dfrac{1}{3}}}-{{\left( 4+3h \right)}^{\dfrac{1}{2}}} \right\}\sin \pi h} \\
\end{align}$
Let us multiply and divide the denominator by $\pi h$ .
$\Rightarrow \displaystyle \lim_{h\to 0}\dfrac{\log \left( 1+\dfrac{h}{2} \right)\left( 3\cdot {{4}^{h}}-3-3h \right)}{-\left\{ {{\left( 8+h \right)}^{\dfrac{1}{3}}}-{{\left( 4+3h \right)}^{\dfrac{1}{2}}} \right\}\dfrac{\sin \pi h}{\pi h}\times \pi h}$
We have to multiply and divide the denominator by 2.
\[\Rightarrow \displaystyle \lim_{h\to 0}\dfrac{\log \left( 1+\dfrac{h}{2} \right)\left( 3\cdot {{4}^{h}}-3-3h \right)}{-\left\{ {{\left( 8+h \right)}^{\dfrac{1}{3}}}-{{\left( 4+3h \right)}^{\dfrac{1}{2}}} \right\}\dfrac{\sin \pi h}{\pi h}\times \dfrac{\pi h}{2}\times 2}\]
We know that $\displaystyle \lim_{x \to a}\dfrac{f\left( x \right)}{g\left( x \right)}=\dfrac{\displaystyle \lim_{x \to a}f\left( x \right)}{\displaystyle \lim_{x \to a}g\left( x \right)}$ and $\displaystyle \lim_{x \to a}f\left( x \right)g\left( x \right)=\displaystyle \lim_{x \to a}f\left( x \right)\times \displaystyle \lim_{x \to a}g\left( x \right)$ .
$\Rightarrow \dfrac{\displaystyle \lim_{h\to 0}\dfrac{\log \left( 1+\dfrac{h}{2} \right)}{\dfrac{h}{2}}}{\displaystyle \lim_{h\to 0}\dfrac{\sin \pi h}{\pi h}}\displaystyle \lim_{h\to 0}\dfrac{\left( 3\cdot {{4}^{h}}-3-3h \right)}{-\left\{ {{\left( 8+h \right)}^{\dfrac{1}{3}}}-{{\left( 4+3h \right)}^{\dfrac{1}{2}}} \right\}2\pi }$
We know that $\displaystyle \lim_{x \to 0}\dfrac{\log \left( 1+x \right)}{x}=1$ and $\displaystyle \lim_{x \to 0}\dfrac{\sin x}{x}=1$ .
$\begin{align}
& \Rightarrow 1\times \displaystyle \lim_{h\to 0}\dfrac{\left( 3\cdot {{4}^{h}}-3-3h \right)}{-\left\{ {{\left( 8+h \right)}^{\dfrac{1}{3}}}-{{\left( 4+3h \right)}^{\dfrac{1}{2}}} \right\}2\pi } \\
& \Rightarrow \displaystyle \lim_{h\to 0}\dfrac{\left( 3\cdot {{4}^{h}}-3-3h \right)}{-\left\{ {{\left( 8+h \right)}^{\dfrac{1}{3}}}-{{\left( 4+3h \right)}^{\dfrac{1}{2}}} \right\}2\pi } \\
\end{align}$
We have to take the common factor 3 outside from the numerator.
$\Rightarrow \displaystyle \lim_{h\to 0}\dfrac{3\left( {{4}^{h}}-1-h \right)}{-\left\{ {{\left( 8+h \right)}^{\dfrac{1}{3}}}-{{\left( 4+3h \right)}^{\dfrac{1}{2}}} \right\}2\pi }$
We can see that when we apply the limit, we will get the result in $\dfrac{0}{0}$ . Therefore, we have to apply L'Hospital's Rule. According to this rule, we will find the derivative of the numerator and the denominator.
$\Rightarrow \displaystyle \lim_{h\to 0}\dfrac{\dfrac{d}{dh}\left\{ 3\left( {{4}^{h}}-1-h \right) \right\}}{\dfrac{d}{dh}\left[ -2\pi \left\{ {{\left( 8+h \right)}^{\dfrac{1}{3}}}-{{\left( 4+3h \right)}^{\dfrac{1}{2}}} \right\} \right]}$
We know that $\dfrac{d}{dx}ax=a\dfrac{d}{dx}x$ .
$\begin{align}
& \Rightarrow \displaystyle \lim_{h\to 0}\dfrac{3\dfrac{d}{dh}\left( {{4}^{h}}-1-h \right)}{-2\pi \dfrac{d}{dh}\left\{ {{\left( 8+h \right)}^{\dfrac{1}{3}}}-{{\left( 4+3h \right)}^{\dfrac{1}{2}}} \right\}} \\
& \Rightarrow \displaystyle \lim_{h\to 0}\dfrac{3\left[ \dfrac{d}{dh}{{4}^{h}}-\dfrac{d}{dh}1-\dfrac{d}{dh}h \right]}{-2\pi \left[ \dfrac{d}{dh}{{\left( 8+h \right)}^{\dfrac{1}{3}}}-\dfrac{d}{dh}{{\left( 4+3h \right)}^{\dfrac{1}{2}}} \right]}...\left( i \right) \\
\end{align}$
Let us consider $y={{4}^{h}}...\left( iii \right)$ . We have to take logarithms on both sides.
$\Rightarrow \log y=\log {{4}^{h}}$
We know that $\log {{x}^{a}}=a\log x$ .
$\Rightarrow \log y=h\log 4$
Let us differentiate both the sides with respect to h. We know that $\dfrac{d}{dx}{{\log }_{e}}x=\dfrac{1}{x}$ and $\dfrac{d}{dx}ax=a\dfrac{d}{dx}x$ . Therefore, the above equation becomes
$\begin{align}
& \Rightarrow \dfrac{1}{y}\dfrac{dy}{dh}={{\log }_{e}}4 \\
& \Rightarrow \dfrac{dy}{dh}=y{{\log }_{e}}4 \\
\end{align}$
We have to substitute (iii) in the above equation.
$\Rightarrow \dfrac{dy}{dh}={{4}^{h}}{{\log }_{e}}4$
Let us substitute the above value in the equation (i).
$\Rightarrow \displaystyle \lim_{h\to 0}\dfrac{3\left[ {{4}^{h}}{{\log }_{e}}4-\dfrac{d}{dh}1-\dfrac{d}{dh}h \right]}{-2\pi \left[ \dfrac{d}{dh}{{\left( 8+h \right)}^{\dfrac{1}{3}}}-\dfrac{d}{dh}{{\left( 4+3h \right)}^{\dfrac{1}{2}}} \right]}$
We know that $\dfrac{d}{dx}a=0$ and $\dfrac{d}{dx}{{x}^{n}}=n{{x}^{n-1}}$ . Therefore, the above equation becomes
$\begin{align}
& \Rightarrow \displaystyle \lim_{h\to 0}\dfrac{3\left[ {{4}^{h}}{{\log }_{e}}4-0-1 \right]}{-2\pi \left[ \dfrac{1}{3}{{\left( 8+h \right)}^{\dfrac{1}{3}-1}}-\dfrac{1}{2}{{\left( 4+3h \right)}^{\dfrac{1}{2}-1}}\times 3 \right]} \\
& \Rightarrow \displaystyle \lim_{h\to 0}\dfrac{3\left[ {{4}^{h}}{{\log }_{e}}4-1 \right]}{-2\pi \left[ \dfrac{1}{3}{{\left( 8+h \right)}^{-\dfrac{2}{3}}}-\dfrac{3}{2}{{\left( 4+3h \right)}^{-\dfrac{1}{2}}} \right]} \\
\end{align}$
Now, let us apply the limit.
$\begin{align}
& \Rightarrow \dfrac{3\left[ {{4}^{0}}{{\log }_{e}}4-1 \right]}{-2\pi \left[ \dfrac{1}{3}{{\left( 8+0 \right)}^{-\dfrac{2}{3}}}-\dfrac{3}{2}{{\left( 4+0 \right)}^{-\dfrac{1}{2}}} \right]} \\
& \Rightarrow \dfrac{3\left[ \log {{4}_{e}}-1 \right]}{-2\pi \left[ \dfrac{1}{3}\cdot {{8}^{-\dfrac{2}{3}}}-\dfrac{3}{2}\cdot {{4}^{-\dfrac{1}{2}}} \right]}...\left( ii \right) \\
\end{align}$
Let us evaluate ${{8}^{-\dfrac{2}{3}}}$ .
$\Rightarrow {{8}^{-\dfrac{2}{3}}}=\dfrac{1}{{{8}^{\dfrac{2}{3}}}}$
We know that ${{\left( {{a}^{m}} \right)}^{n}}={{a}^{mn}}$ .
$\Rightarrow {{8}^{-\dfrac{2}{3}}}=\dfrac{1}{{{\left( {{8}^{2}} \right)}^{\dfrac{1}{3}}}}$
$\Rightarrow {{8}^{-\dfrac{2}{3}}}=\dfrac{1}{\sqrt[3]{{{8}^{2}}}}$
$\Rightarrow {{8}^{-\dfrac{2}{3}}}=\dfrac{1}{\sqrt[3]{64}}$
$\Rightarrow {{8}^{-\dfrac{2}{3}}}=\dfrac{1}{4}...\left( iii \right)$
Now, let us find ${{4}^{-\dfrac{1}{2}}}$ .
$\Rightarrow {{4}^{-\dfrac{1}{2}}}=\dfrac{1}{\sqrt{4}}=\dfrac{1}{2}...\left( iv \right)$
We have to substitute (iii) and (iv) in equation (ii).
$\begin{align}
& \Rightarrow \dfrac{3\left[ {{\log }_{e}}4-1 \right]}{-2\pi \left[ \dfrac{1}{3}\cdot \dfrac{1}{4}-\dfrac{3}{2}\cdot \dfrac{1}{2} \right]} \\
& =\dfrac{3\left[ {{\log }_{e}}4-1 \right]}{-2\pi \left[ \dfrac{1}{12}-\dfrac{3}{4} \right]} \\
& =\dfrac{3\left[ {{\log }_{e}}4-1 \right]}{-2\pi \times \dfrac{-8}{12}} \\
& =\dfrac{3\left[ {{\log }_{e}}4-1 \right]}{-2\pi \times \dfrac{-2}{3}} \\
\end{align}$
Let us simplify the above expression.
$\begin{align}
& \Rightarrow \dfrac{3\left[ {{\log }_{e}}4-1 \right]\times 3}{4\pi } \\
& =\dfrac{9\left[ {{\log }_{e}}4-1 \right]}{4\pi } \\
\end{align}$
We know that ${{\log }_{e}}e=1$ . Therefore, we can write the above equation as
$\Rightarrow \dfrac{9\left[ {{\log }_{e}}4-{{\log }_{e}}e \right]}{4\pi }$
We know that $\log a-\log b=\log \left( \dfrac{a}{b} \right)$ . Therefore, we can write the above equation as
\[\begin{align}
& \Rightarrow \dfrac{9{{\log }_{e}}\left( \dfrac{4}{e} \right)}{4\pi } \\
& =\dfrac{9}{4\pi }{{\log }_{e}}\left( \dfrac{4}{e} \right) \\
\end{align}\]
Hence, the value of $\displaystyle \lim_{x \to 1}\dfrac{\left( \log \left( 1+x \right)-\log 2 \right)\left( 3\cdot {{4}^{x-1}}-3x \right)}{\left\{ {{\left( 7+x \right)}^{\dfrac{1}{3}}}-{{\left( 1+3x \right)}^{\dfrac{1}{2}}} \right\}\sin \pi x}$ is \[\dfrac{9}{4\pi }{{\log }_{e}}\left( \dfrac{4}{e} \right)\] .
Note: Students must be thorough with logarithmic rules and properties. They must deeply learn rules associated with limits and how to evaluate limits using L-Hospitals rule. Students must know to find the derivatives.
Recently Updated Pages
Master Class 11 Computer Science: Engaging Questions & Answers for Success
Master Class 11 Business Studies: Engaging Questions & Answers for Success
Master Class 11 Economics: Engaging Questions & Answers for Success
Master Class 11 English: Engaging Questions & Answers for Success
Master Class 11 Maths: Engaging Questions & Answers for Success
Master Class 11 Biology: Engaging Questions & Answers for Success
Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE
There are 720 permutations of the digits 1 2 3 4 5 class 11 maths CBSE
Discuss the various forms of bacteria class 11 biology CBSE
Draw a diagram of a plant cell and label at least eight class 11 biology CBSE
State the laws of reflection of light
Explain zero factorial class 11 maths CBSE