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  • Using the method of L’ Hospital, evaluate the limit. <img alt="\lim _{x \rightarrow \infty}\left[\frac{(\sqrt[x]{64})^{\frac{1}{2}}-1}{\log _c\left(1+\frac{8}{x}

Using the method of L’ Hospital, evaluate the limit.

\lim _{x \rightarrow \infty}\left[\frac{(\sqrt[x]{64})^{\frac{1}{2}}-1}{\log _c\left(1+\frac{8}{x}\right)}\right]

Option: 1

\frac{\ln 8}{8}


Option: 2

In2


Option: 3

\frac{\ln 2}{8}

 


Option: 4

\frac{\ln 2}{2}


Answers (1)

best_answer

Apply the L' Hospital's Rule, when   \lim _{x \rightarrow 0} \frac{f(x)}{g(x)}=\frac{0}{0}, \frac{\infty}{\infty}, \text { etc. }   [an intermediate form] in the following way

  • Differentiate \lim _{x \rightarrow a} \frac{f(x)}{g(x)}=\lim _{x \rightarrow a} \frac{f^{\prime}(x)}{g^{\prime}(x)}, provided that the limit \lim _{x \rightarrow a} \frac{f^{\prime}(x)}{g^{\prime}(x)} exists.

  • Otherwise, go on differentiating till the limit with the determinate form is achieved.

Assume  x\frac{1}{p}

So, as x \rightarrow \infty , p\rightarrow 0

The provided limit is

\begin{aligned} & \lim _{x \rightarrow \infty}\left[\frac{(\sqrt[x]{64})^{\frac{1}{2}}-1}{\log _e\left(1+\frac{8}{x}\right)}\right] \\ & =\lim _{x \rightarrow \infty}\left[\frac{8^{\frac{1}{x}}-1}{\log _e\left(1+\frac{8}{x}\right)}\right] \quad\left[\because(\sqrt[x]{64})^{\frac{1}{2}}=64^{\frac{1}{2 x}}=8^{\left(2 \frac{1}{2 x}\right)}=8^{\frac{1}{x}}\right] \\ & =\lim _{p \rightarrow 0}\left[\frac{\left(8^p-1\right)}{\log _e(1+8 p)}\right]=\frac{0}{0} \end{aligned}

Here use the L’ Hospital rule and also refer to the following formula.

\begin{aligned} & \frac{d}{d x}\left(e^{a x}\right)=a e^{a x} \\ & \frac{d}{d x}\left(\log _e x\right)=\frac{1}{x} \end{aligned}

Now, derive the following:

\begin{aligned} & \lim _{p \rightarrow 0}\left[\frac{\left(8^p-1\right)}{\log _e(1+8 p)}\right] \\ & =\lim _{p \rightarrow 0}\left[\frac{\frac{d}{d p}\left(8^p-1\right)}{d \log _e(1+8 p)}\right] \\ & =\lim _{p \rightarrow 0}\left[\frac{8^p \ln 8}{\frac{8}{(1+8 p)}}\right] \quad\left[y \frac{d}{d x}\left(b^x\right)=b^x \ln b, \frac{d}{d x}\left(\log _e x\right)=\frac{1}{x}, \frac{d}{d x} x^n=n x^{n 1}\right] \end{aligned}

\begin{aligned} & =\frac{8^0 \times \ln 8}{\frac{8}{(1+0)}} \\ & =\frac{\ln 8}{8} \end{aligned}

 

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Gunjita

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