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Limits, Continuity and Differentiability Part 1

Section: Mathematics

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Question : 3 of 100

Marks: +1, -0

\lim\limits_{x\to 0}\frac{x\left(e^{\frac{\sqrt{1+x^{2}+x^{4}}-1}{x}}-1\right)}{\sqrt{1+x^{2}+x^{4}}-1}

[5 Sep 2020 Shift 2]

does not exist.
is equal to $\sqrt{{e}}$ .
is equal to 0 .
is equal to 1.

Solution:

\lim\limits_{x\to 0}\frac{x\left(e^{\frac{\sqrt{1+x^{2}+x^{4}}-1}{x}}-1\right)}{\sqrt{1+x^{2}+x^{4}}-1}

\because\lim\limits_{x\to 0}(\sqrt{1+x^{2}+x^{4}}-1)x\left(\frac{0}{0},\right)

from

\lim\limits_{x\to 0}\frac{(1+x^{2}+x^{4})-1}{x(\sqrt{1+x^{2}+x^{4}}+1)}

\lim\limits_{x\to 0}\frac{x(1+x^{2})}{\sqrt{1+x^{2}+x^{4}}+1}=0

\lim\limits_{x\to 0}\frac{x\left(e^{\frac{\sqrt{1+x^{2}+x^{4}}-1}{x}}-1\right)}{\sqrt{1+x^{2}+x^{4}}-1}

So

\;\left(\frac{0}{0},\right)

\lim\limits_{x\to 0}\frac{e^{(\sqrt{1+x^{2}+x^{4}}-1)x}-1}{\frac{\sqrt{1+x^{2}+x^{4}}-1}{x}}=1

from

\sqrt{e}

\lim\limits_{x\to 0}\frac{e^{\frac{\sqrt{1+x^{2}+x^{4}}-1}{x}}-1}{\frac{\sqrt{1+x^{2}+x^{4}}-1}{x}}=1

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