Evaluate

Question:

Evaluate

$\lim _{x \rightarrow 2}\left(\frac{3^{x}-3^{3-x}-12}{3^{3-x}-3^{x / 2}}\right)$

 

Solution:

To evaluate:

$\lim _{x \rightarrow 2}\left(\frac{3^{x}-3^{3-x}-12}{3^{3-x}-3^{\frac{x}{2}}}\right)$

Formula used:

L'Hospital's rule

Let $f(x)$ and $g(x)$ be two functions which are differentiable on an open interval I except at a point a where

then

$\lim _{x \rightarrow a} \frac{\mathrm{f}(\mathrm{x})}{\mathrm{g}(\mathrm{x})}=\lim _{x \rightarrow a} \frac{\mathrm{f}^{\prime}(\mathrm{x})}{\mathrm{g}^{\prime}(\mathrm{x})}$

As $\mathrm{X} \rightarrow 0$, we have

$\lim _{x \rightarrow 0} \frac{2^{x}-1}{x}=\frac{0}{0}$

This represents an indeterminate form. Thus applying L'Hospital's rule, we get

$\lim _{x \rightarrow 2}\left(\frac{3^{x}-3^{3-x}-12}{3^{3-x}-3^{\frac{x}{2}}}\right)=\lim _{x \rightarrow 2} \frac{\frac{d}{d x}\left(3^{x}-3^{3-x}-12\right)}{\frac{d}{d x}\left(3^{3-x}-3^{\frac{x}{2}}\right)}$

$\lim _{x \rightarrow 2}\left(\frac{3^{x}-3^{3-x}-12}{3^{3-x}-3^{\frac{x}{2}}}\right)=\lim _{x \rightarrow 0} \frac{3^{x} \ln 3+3^{3-x} \ln 3}{-3^{3-x} \ln 3+3^{\frac{x}{2}}\left(\frac{1}{2}\right) \ln 3}$

$\lim _{x \rightarrow 2}\left(\frac{3^{x}-3^{3-x}-12}{3^{3-x}-3^{\frac{x}{2}}}\right)=\frac{\ln 3+27 \ln 3}{-27 \ln 3+\left(\frac{1}{2}\right) \ln 3}$

$\lim _{x \rightarrow 2}\left(\frac{3^{x}-3^{3-x}-12}{3^{3-x}-3^{\frac{x}{2}}}\right)=\frac{28 \ln 3}{-26.5 \ln 3}$

Thus, the value of $\lim _{x \rightarrow 2}\left(\frac{3^{x}-3^{3-x}-12}{3^{3-x}-3^{\frac{x}{2}}}\right)$ is $\frac{28 \ln 3}{-26.5 \ln 3}$

 

 

 

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