Let f(x)=x2+1−x2+13×4+14−x4+15Statement-1 limx→∞ f(x)=1Statement-2 limx→+∞ 1xn=0 for n>0

# Let $f\left(x\right)=\frac{\sqrt{{x}^{2}+1}-\sqrt[3]{{x}^{2}+1}}{\sqrt[4]{{x}^{4}+1}-\sqrt[5]{{x}^{4}+1}}$Statement-$1$ $\underset{x\to \mathrm{\infty }}{lim} f\left(x\right)=1$Statement-$2$ $\underset{x\to +\mathrm{\infty }}{lim} \frac{1}{{x}^{n}}=0$ for $n>0$

1. A

STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is a correct explanation for STATEMENT-1

2. B

STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is NOT a correct explanation for
STATEMENT-1

3. C

STATEMENT-1 is True, STATEMENT-2 is False

4. D

STATEMENT-1 is False, STATEMENT-2 is True

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### Solution:

$f\left(x\right)=\frac{x{\left(1+1/{x}^{2}\right)}^{1/2}-{x}^{2/3}{\left(1+1/{x}^{2}\right)}^{1/3}}{x{\left(1+1/{x}^{4}\right)}^{1/4}-{x}^{4/5}{\left(1+1/{x}^{4}\right)}^{1/5}}$

$=\frac{\left(1+\frac{1}{2}\frac{1}{{x}^{2}}+0\left(\frac{1}{{x}^{2}}\right)\right)-\frac{1}{{x}^{1/3}}\left(1+\frac{1}{3{x}^{2}}+0\left(\frac{1}{{x}^{2}}\right)\right)}{\left(1+\frac{1}{4}\frac{1}{{x}^{4}}+0\left(\frac{1}{{x}^{4}}\right)\right)-\frac{1}{{x}^{1/5}}\left(1+\frac{1}{5{x}^{4}}+0\left(\frac{1}{{x}^{4}}\right)\right)}$

Since $\underset{x\to \mathrm{\infty }}{lim} \frac{1}{{x}^{n}}=0$ for all $n>0$ so, $\underset{x\to \mathrm{\infty }}{lim} f\left(x\right)=\frac{1-0.1}{1-0.1}=1$