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  • Using the method of L' Hospital, evaluate the limit. <img alt="\lim _{x \rightarrow 0} \frac{\left(25^{x}\right)^{5}-\left(\sqrt{25^{x}}\right)^{5}}{x}" src="https://entrancecorner.oncodec

Using the method of L' Hospital, evaluate the limit. \lim _{x \rightarrow 0} \frac{\left(25^{x}\right)^{5}-\left(\sqrt{25^{x}}\right)^{5}}{x}

Option: 1

\ln 25


Option: 2

\ln 5


Option: 3

5\ln 5


Option: 4

5\ln 10


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}, 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.

The provided limit is

\lim _{x \rightarrow 0} \frac{\left(25^{x}\right)^{5}-\left(\sqrt{25^{x}}\right)^{5}}{x}=\frac{0}{0}

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

\frac{d}{d x}\left(b^{x}\right)=b^{x} \ln b

Now, derive the following: 

\begin{aligned} & \lim _{x \rightarrow 0} \frac{\left(25^{x}\right)^{5}-\left(\sqrt{25^{x}}\right)^{5}}{x} \\ & =\lim _{x \rightarrow 0} \frac{25^{5 x}-25^{\frac{5 x}{2}}}{x} \\ & =\lim _{x \rightarrow 0} \frac{\frac{d}{d x}\left(25^{5 x}-25^{\frac{5 x}{2}}\right)}{\frac{d}{d x}(x)} \\ & =\lim _{x \rightarrow 0} \frac{5 \times 25^{5 x} \ln 25-\frac{5}{2} \times 25^{\frac{5 x}{2}} \ln 25}{1} \quad\left[ \frac{d}{d x}\left(b^{x}\right)=b^{x} \ln b\right] \\ \end{aligned}\begin{aligned} & =5 \times 1 \times \ln 25-\frac{5}{2} \times 1 \times \ln 25 \\ & =\frac{5}{2} \ln 25 \\ & =\frac{5}{2} \ln 5^{2} \\ & =\frac{5}{2} \times 2 \ln 5 \\ & =5 \ln 5 \end{aligned}

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