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Without using the method of L’ Hospital, find the limit.

\lim _{x \rightarrow \infty} x\left[(\sqrt[x]{16})^{\frac{1}{4}}+\log _e\left(1+\frac{12}{x}\right)-1\right]

Option: 1

\ln 2+12


Option: 2

\ln 2 \pm 12


Option: 3

\ln 2-12


Option: 4

Cannot be determined


Answers (1)

best_answer

Note the following exponential limit.

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

\frac{\log _e(1+x)}{x}=1 \\

\lim _{x \rightarrow 0} \frac{a^x-1}{x}=\log _c a=\ln a \\

Assume   x=\frac{1}{p}

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

Now, evaluate the following limit using the above formula and the assumption.

\begin{aligned} & \lim _{x \rightarrow x} x\left[(\sqrt{16})^{\frac{1}{4}}+\log _e\left(1+\frac{12}{x}\right)-1\right] \\ & =\lim _{x \rightarrow \infty} x\left[\left(\sqrt[x]{2^4}\right)^{\frac{1}{4}}+\log _e\left(1+\frac{12}{x}\right)-1\right] \\ & =\lim _{x \rightarrow \infty}\left[x\left(2^{\frac{4}{4 x}}-1\right)+x \log _a\left(1+\frac{12}{x}\right)\right] \\ & =\lim _{x \rightarrow \infty}\left[x\left(2^x-1\right)+x \log _c\left(1+\frac{12}{x}\right)\right] \\ & =\lim _{p \rightarrow 0}\left[\frac{1}{p}\left(2^p-1\right)+\frac{1}{p} \log _e(1+12 p)\right] \\ & =\lim _{p \rightarrow 0} \frac{\left(2^p-1\right)}{p}+12 \times \frac{\log _e(1+12 p)}{12 p} \\ & =\ln ^2 2+12 \times 1 \\ & =\ln ^2 2+12 \end{aligned}

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shivangi.shekhar

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