Let f(x)=1+sin⁡x−1−sin⁡xtan⁡x,x≠0and g(x)=1+1xx+1xStatement-1: limx→0 f(x)=limx→∞ g(x)Statement-2: Both the limits are equal to 1.

# Let $f\left(x\right)=\frac{\sqrt{1+\mathrm{sin}x}-\sqrt{1-\mathrm{sin}x}}{\mathrm{tan}x},x\ne 0$and $g\left(x\right)={\left(1+\frac{1}{x}\right)}^{\frac{x+1}{x}}$Statement-$1:$ $\underset{x\to 0}{lim} f\left(x\right)=\underset{x\to \mathrm{\infty }}{lim} g\left(x\right)$Statement-$2:$ Both the limits are equal to

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 Tru

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### Solution:

$\underset{x\to 0}{lim} f\left(x\right)$

$\begin{array}{l}=\underset{x\to 0}{lim} \frac{1+\mathrm{sin}x-1+\mathrm{sin}x}{\mathrm{tan}x}×\frac{1}{\sqrt{1+\mathrm{sin}x}+\sqrt{1-\mathrm{sin}x}}\\ =\underset{x\to 0}{lim} 2\mathrm{cos}x×\frac{1}{\sqrt{1+\mathrm{sin}x}+\sqrt{1-\mathrm{sin}x}}\\ =2×1×\frac{1}{1+1}=1.\\ \begin{array}{l}\underset{x\to \mathrm{\infty }}{lim} g\left(x\right)=\underset{x\to \mathrm{\infty }}{lim} {\left(1+\frac{1}{x}\right)}^{x×\left(\frac{1}{x}+\frac{1}{{x}^{2}}\right)}={e}^{0}=1\end{array}\end{array}$

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