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 $L=\underset{x\to 0}{\mathop{\lim }}\,\dfrac{{{2}^{3x}}-{{3}^{x}}}{\sin 3x}$.

& \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{aligned}$$ If a limit of a function $$\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.