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  • Without using the method of L' Hospital, find the limit.  <img alt="\lim _{x \rightarrow \infty} x^{2}\left[\left((\sqrt[x]{16})^{\frac{1}{4}}-1\right) \log _{e}\left(1+\frac{12}

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

\lim _{x \rightarrow \infty} x^{2}\left[\left((\sqrt[x]{16})^{\frac{1}{4}}-1\right) \log _{e}\left(1+\frac{12}{x}\right)\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.

$$ \begin{gathered} \cdot \quad \lim _{x \rightarrow 0} \frac{e^{x}-1}{x}=1 \\ \cdot \quad \lim _{x \rightarrow 0} \frac{\log _{e}(1+x)}{x}=1 \\ \cdot \quad \lim _{x \rightarrow 0} \frac{a^{x}-1}{x}=\log _{e} a=\ln a \\ \end{gathered}

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

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Ritika Harsh

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