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# Evaluate $\underset{x\to 1}{\mathop{\lim }}\,\dfrac{\left( \ln \left( 1+x \right)-\ln 2 \right)\left( {{3.4}^{x-1}}-3x \right)}{\left[ {{\left( 7+x \right)}^{\dfrac{1}{3}}}-{{\left( 1+3x \right)}^{\dfrac{1}{2}}} \right]\sin \left( x-1 \right)}$.

Last updated date: 14th Jul 2024
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Hint: Given limit $x\to 1$, make the limit $h\to 0$. Simplify the expression and substitute using the L-Hospital rule if you get indeterminate form at h=0 as $\dfrac{0}{0}$. After simplifying the expression put, $h\to 0$and evaluate the expression.

We have to evaluate the given limit which is $x\to 1$.
Let us put, $x=1+h$.
If x = 1, then, $1=1+h\Rightarrow h=0$.
Therefore, limit $x\to 1$changes to limit $h\to 0$.
Therefore, to evaluate changes, put $x=1+h$.
\begin{align} & \underset{h\to 0}{\mathop{\lim }}\,\dfrac{\left( \ln \left( 1+1+h \right)-\ln 2 \right)\left( {{3.4}^{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 \left[ 1+h-1 \right]} \\ & =\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\left( \ln \left( 2+h \right)-\ln 2 \right)\left( {{3.4}^{h}}-3-3h \right)}{\left[ {{\left( 8+h \right)}^{\dfrac{1}{3}}}-{{\left( 4+3h \right)}^{\dfrac{1}{2}}} \right]\sinh } \\ \end{align}
We know, $\ln a-\ln b=\ln \left( \dfrac{a}{b} \right)$.
$\therefore \ln \left( 2+h \right)-\ln 2=\ln \left( \dfrac{2+h}{2} \right)=\ln \left( 1+\dfrac{h}{2} \right)$
$=\lim \dfrac{\ln \left( 1+\dfrac{h}{2} \right)\left( {{3.4}^{h}}-3-3h \right)}{\left[ {{\left( 8+h \right)}^{\dfrac{1}{3}}}-{{\left( 4+3h \right)}^{\dfrac{1}{2}}} \right]\sinh }$
$=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\ln \left( 1+\dfrac{h}{2} \right)\left[ 3\left( {{4}^{h}}-1-h \right) \right]}{\left[ {{\left( 8+h \right)}^{\dfrac{1}{3}}}-{{\left( 4+3h \right)}^{\dfrac{1}{2}}} \right]\sinh }$
We know, $\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\sinh }{h}=1$.
Similarly, $\underset{a\to 0}{\mathop{\lim }}\,\dfrac{\ln \left( 1+a \right)}{a}=1$.
Let us multiply 2h in the numerator and denominator.
\begin{align} & =\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\ln \left( 1+\dfrac{h}{2} \right)\left[ 3\left( {{4}^{h}}-1-h \right) \right]\times 2\times h}{\left[ {{\left( 8+h \right)}^{\dfrac{1}{3}}}-{{\left( 4+3h \right)}^{\dfrac{1}{2}}} \right]\sinh \times 2\times h} \\ & =\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\ln \left( 1+\dfrac{h}{2} \right)\left[ 3\left( {{4}^{h}}-1-h \right) \right]}{\left[ {{\left( 8+h \right)}^{\dfrac{1}{3}}}-{{\left( 4+3h \right)}^{\dfrac{1}{2}}} \right]\sinh \times \dfrac{2h}{2h}} \\ & =\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\ln \left( 1+\dfrac{h}{2} \right)\left[ 3\left( {{4}^{h}}-1-h \right) \right]}{\dfrac{h}{2}\left[ {{\left( 8+h \right)}^{\dfrac{1}{3}}}-{{\left( 4+3h \right)}^{\dfrac{1}{2}}} \right]\left( \dfrac{\sinh }{h} \right)\times 2} \\ & =\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\ln \left( 1+\dfrac{h}{2} \right)\left[ 3\left( {{4}^{h}}-1-h \right) \right]}{\dfrac{h}{2}\left[ {{\left( 8+h \right)}^{\dfrac{1}{3}}}-{{\left( 4+3h \right)}^{\dfrac{1}{2}}} \right]\left[ \dfrac{\sinh }{h} \right]\times 2} \\ \end{align}
$\dfrac{\ln \left( 1+\dfrac{h}{2} \right)}{\dfrac{h}{2}}=1$and $\dfrac{\sinh }{h}=1$.
L-Hospital rule states that for functions f and g are differentiable on an open interval I except possibly at point C contained I, if $\underset{x\to c}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to c}{\mathop{\lim }}\,g\left( x \right)=0$or $\pm \infty$, $g'\left( x \right)\ne 0$for all x in I with $x\ne c$, thus $\underset{x\to c}{\mathop{\lim }}\,\dfrac{f\left( x \right)}{g\left( x \right)}=\underset{x\to c}{\mathop{\lim }}\,\dfrac{f'\left( x \right)}{g'\left( x \right)}$.
If we apply $h\to 0$in equation (1), then the indeterminate form at $h\to 0$is $\dfrac{0}{0}$.
Thus, differentiate the numerator and denominator to simplify the expression to a limit that can be evaluated directly.
Let us put, $y={{4}^{h}}$.
\begin{align} & \ln y=\ln \left( {{4}^{h}} \right) \\ & \ln y=h\ln 4 \\ \end{align}.
Differentiating both sides,
$\dfrac{1}{y}\dfrac{dy}{dh}=\ln 4$
$\Rightarrow \dfrac{dy}{dh}=y\ln 4$ (where, $y={{4}^{h}}$)
$\Rightarrow \dfrac{dy}{dh}={{4}^{h}}.\ln 4$
Applying L-Hospital rule once,
$=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{3\left( {{4}^{h}}.\ln 4-1 \right)}{2\left[ \dfrac{1}{3}{{\left( 8+h \right)}^{\dfrac{1}{3}-1}}-\dfrac{3}{2}{{\left( 4+3h \right)}^{\dfrac{1}{2}-1}} \right]}$
Put h = 0, in the above equation.
\begin{align} & =\dfrac{3\left( {{4}^{h}}.\ln 4-1 \right)}{2\left[ \dfrac{1}{3}{{8}^{\dfrac{-2}{3}}}-\dfrac{3}{2}{{4}^{\dfrac{-1}{2}}} \right]}=\dfrac{3\left( \ln 1-1 \right)}{2\left[ \dfrac{1}{3}\times \dfrac{1}{{{8}^{\dfrac{2}{3}}}}-\dfrac{1}{2}\times \dfrac{3}{{{4}^{\dfrac{1}{2}}}} \right]} \\ & =\dfrac{3\left( \ln 1-1 \right)}{2\left[ \dfrac{1}{3}\times \dfrac{1}{{{\left( {{2}^{3}} \right)}^{\dfrac{2}{3}}}}-\dfrac{1}{2}\times \dfrac{3}{2} \right]} \\ & =\dfrac{3\left( \ln 1-1 \right)}{2\left[ \dfrac{1}{3}\times \dfrac{1}{{{2}^{2}}}-\dfrac{3}{4} \right]}=\dfrac{3\left( \ln 1-1 \right)}{2\left[ \dfrac{1}{12}-\dfrac{3}{4} \right]} \\ & =\dfrac{3\left( 0-1 \right)}{2\left[ \dfrac{4-36}{12\times 4} \right]}=\dfrac{-3}{\dfrac{-2\times \left( 32 \right)}{12\times 4}}=\dfrac{-3}{\dfrac{-64}{48}}=\dfrac{-9}{4}\left( -1 \right)=\dfrac{9}{4} \\ \end{align}
Hence, the limit evaluates to $\dfrac{9}{4}$.

Note: In limits be careful to use formula to simplify the expression wherever necessary. We have used substitutions and formulae to solve the limit. Remember these steps involved and how to change the limit $\left( x\to 1 \right)$to $h\to 0$. You can’t apply $\left( x\to 1 \right)$in an expression like this. So remember to convert it to $h\to 0$.