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When \lim _{x \rightarrow a} f(x)=0, which of the following is true?

Option: 1

\lim _{x \rightarrow a} \frac{\log _{e}(1+f(x))}{f(x)} \times \lim _{x \rightarrow a} \frac{a^{f(x)}-1}{f(x)}=1


Option: 2

\lim _{x \rightarrow a} \frac{a^{f(x)}-1}{f(x)} \times \times \lim _{x \rightarrow a} \frac{e^{f(x)}-1}{f(x)}=1


Option: 3

\lim _{x \rightarrow a} \frac{\log _{e}(1+f(x))}{f(x)} \times \lim _{x \rightarrow a} \frac{e^{f(x)}-1}{f(x)}=1


Option: 4

 None of the above


Answers (1)

best_answer

For the limit \lim _{x \rightarrow a} f(x)=0, the following hold well.

\begin{aligned} & \text { . } \lim _{x \rightarrow a} \frac{\log _{e}(1+f(x))}{f(x)}=1 \\ & \text { . } \lim _{x \rightarrow a} \frac{a^{f(x)}-1}{f(x)}=\log _{e} a \\ & \text { . } \lim _{x \rightarrow a} \frac{e^{f(x)}-1}{f(x)}=\log _{e} e=1 \end{aligned}

From the above limits, the following is evident.

\begin{aligned} & \lim _{x \rightarrow a} \frac{\log _{e}(1+f(x))}{f(x)} \times \lim _{x \rightarrow a} \frac{e^{f(x)}-1}{f(x)} \\ & =1 \times 1 \\ & =1 \end{aligned}

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manish painkra

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