Question

Find the limit: \[\underset{x\to 0}{\mathop{\lim }}\,\dfrac{{{2}^{3x}}-{{3}^{x}}}{\sin 3x}\].

Answer 1

Hint: L-Hospital rule is used if the value of limit is equal to \[\dfrac{0}{0}\] (or) \[\dfrac{\infty }{\infty }\]. According to L-hospital rule, if \[\underset{x\to a}{\mathop{\lim }}\,\dfrac{f(x)}{g(x)}\]is equal \[\dfrac{0}{0}\] (or) \[\dfrac{\infty }{\infty }\], then \[\underset{x\to a}{\mathop{\lim }}\,\dfrac{f(x)}{g(x)}=\underset{x\to a}{\mathop{\lim }}\,\dfrac{f’(x)}{g’(x)}\]. For any other conditions, L-hospital should not be used. L-hospital is used to evaluate limits for indeterminate forms.

Complete step-by-step solution -
From the question, it is clear that we needed to find the value of \[\underset{x\to 0}{\mathop{\lim }}\,\dfrac{{{2}^{3x}}-{{3}^{x}}}{\sin 3x}\].
Let us assume \[L=\underset{x\to 0}{\mathop{\lim }}\,\dfrac{{{2}^{3x}}-{{3}^{x}}}{\sin 3x}\].
\[\begin{align}
  & \Rightarrow L=\underset{x\to 0}{\mathop{\lim }}\,\dfrac{{{2}^{3(0)}}-{{3}^{(0)}}}{\sin 3(0)} \\
 & \Rightarrow L=\dfrac{{{2}^{0}}-{{3}^{0}}}{\sin 0} \\
 & \Rightarrow L=\dfrac{1-1}{0} \\
 & \Rightarrow L=\dfrac{0}{0} \\
\end{align}\]
If a limit of a function \[\dfrac{f(x)}{g(x)}\] is \[\dfrac{0}{0}\] (or) \[\dfrac{\infty }{\infty }\], then L-Hospital rule is used.
According to L-hospital rule, if \[\underset{x\to a}{\mathop{\lim }}\,\dfrac{f(x)}{g(x)}\]is equal \[\dfrac{0}{0}\] (or) \[\dfrac{\infty }{\infty }\], then \[\underset{x\to a}{\mathop{\lim }}\,\dfrac{f(x)}{g(x)}=\underset{x\to a}{\mathop{\lim }}\,\dfrac{f’(x)}{g’(x)}\].
As the value of \[\underset{x\to 0}{\mathop{\lim }}\,\dfrac{{{2}^{3x}}-{{3}^{x}}}{\sin 3x}\] is equal to \[\dfrac{0}{0}\], we should use L-Hospital rule.
\[\Rightarrow L=\underset{x\to 0}{\mathop{\lim }}\,\dfrac{{{2}^{3x}}-{{3}^{x}}}{\sin 3x}=\underset{x\to 0}{\mathop{\lim }}\,\dfrac{\dfrac{d}{dx}({{2}^{3x}}-{{3}^{x}})}{\dfrac{d}{dx}(\sin 3x)}\]
\[\Rightarrow L=\underset{x\to 0}{\mathop{\lim }}\,\dfrac{\dfrac{d}{dx}({{2}^{3x}})-\dfrac{d}{dx}({{3}^{x}})}{\dfrac{d}{dx}(\sin 3x)}\]

\[\Rightarrow L=\underset{x\to 0}{\mathop{\lim }}\,\dfrac{\dfrac{d}{dx}({{8}^{x}})-\dfrac{d}{dx}({{3}^{x}})}{\dfrac{d}{dx}(\sin 3x)}\]
We know that \[\dfrac{d}{dx}({{a}^{x}})={{a}^{x}}\log a\] where a is constant.
\[\Rightarrow L=\underset{x\to 0}{\mathop{\lim }}\,\dfrac{({{8}^{x}}\log 8)-({{3}^{x}}log3)}{\dfrac{d}{dx}(\sin 3x)}\]
We know that \[\dfrac{d}{dx}(\sin ax)=a\cos ax\] where a is constant.
\[\Rightarrow L=\underset{x\to 0}{\mathop{\lim }}\,\dfrac{({{8}^{x}}\log 8)-({{3}^{x}}log3)}{3\cos 3x}\]
\[\Rightarrow L=\underset{x\to 0}{\mathop{\lim }}\,\dfrac{({{8}^{x}}\log {{2}^{3}})-({{3}^{x}}log3)}{3\cos 3x}\]
We know that \[{{\operatorname{logx}}^{a}}=a\log x\].
\[\Rightarrow L=\underset{x\to 0}{\mathop{\lim }}\,\dfrac{({{8}^{x}}(3\log 2))-({{3}^{x}}log3)}{3\cos 3x}\]
\[\Rightarrow L=\underset{x\to 0}{\mathop{\lim }}\,\dfrac{{{\left( {{2}^{3}} \right)}^{x}}(3\log 2)-({{3}^{x}}\log 3)}{3\cos 3x}\]
\[\Rightarrow L=\dfrac{{{({{2}^{3}})}^{0}}(3\log 2)-({{3}^{0}}\log 3)}{3\cos 3(0)}\]
\[\Rightarrow L=\dfrac{(3log2-log3)}{3\cos 0}\]
\[\Rightarrow L=\dfrac{\log 8-\log 3}{3(1)}\]
\[\Rightarrow L=\dfrac{\log \left( \dfrac{8}{3} \right)}{3}\]
Hence, the limit of \[\underset{x\to 0}{\mathop{\lim }}\,\dfrac{{{2}^{3x}}-{{3}^{x}}}{\sin 3x}\] is equal to \[\underset{x\to 0}{\mathop{\lim }}\,\dfrac{\log \left( \dfrac{8}{3} \right)}{3}\].

Note: We needed to apply L-hospital if and only if the value of limit is equal \[\dfrac{0}{0}\] (or) \[\dfrac{\infty }{\infty }\]. If the value of limit is obtained in other forms, we should not apply to L-Hospital. If \[\underset{x\to a}{\mathop{\lim }}\,f{{(x)}^{g(x)}}\] is equal to \[{{1}^{\infty }}\], then the value of limit of function is \[\underset{x\to a}{\mathop{\lim }}\,f{{(x)}^{g(x)}}={{e}^{\underset{x\to a}{\mathop{\lim }}\,(f(x)-1)g(x)}}\]. We should be careful while using the formulae of logarithms. We should remember that \[\log (a-b)=\log a-\log b\] is not correct. \[\log (a-b)=\log \left( \dfrac{a}{b} \right)\] is correct. Students may go wrong at this point.