Question:
\( \lim_{x\to0}\frac{x\left(e^{\left(\sqrt{1+x^2+x^4}-1\right)/x}-1\right)}{\sqrt{1+x^2+x^4}-1} \)
\( \lim_{x\to0}\frac{x\left(e^{\left(\sqrt{1+x^2+x^4}-1\right)/x}-1\right)}{\sqrt{1+x^2+x^4}-1} \)
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Put \(h=\frac{\sqrt{1+x^2+x^4}-1}{x}\to0\); the expression becomes \(\frac{e^h-1}{h}\).
Updated On: Nov 5, 2026
- is equal to 0.
- is equal to \( \sqrt{e} \).
- is equal to 1.
- does not exist.
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The Correct Option is C
Solution and Explanation
Let \(h=\frac{\sqrt{1+x^2+x^4}-1}{x}\). Then \(\sqrt{1+x^2+x^4}-1=xh\).
The expression equals \( \frac{x\,(e^h-1)}{xh}=\frac{e^h-1}{h} \).
As \(x\to0\): \(h=\frac{x^2+x^4}{x\left(\sqrt{1+x^2+x^4}+1\right)}=\frac{x(1+x^2)}{\sqrt{1+x^2+x^4}+1}\to0\) (from both sides).
Hence limit \(=\lim_{h\to0}\frac{e^h-1}{h}=1\).
Answer: (3) is equal to 1.
The expression equals \( \frac{x\,(e^h-1)}{xh}=\frac{e^h-1}{h} \).
As \(x\to0\): \(h=\frac{x^2+x^4}{x\left(\sqrt{1+x^2+x^4}+1\right)}=\frac{x(1+x^2)}{\sqrt{1+x^2+x^4}+1}\to0\) (from both sides).
Hence limit \(=\lim_{h\to0}\frac{e^h-1}{h}=1\).
Answer: (3) is equal to 1.
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