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.

limx→0 f(x)=limx→0 1+sin⁡x−1+sin⁡xtan⁡x×11+sin⁡x+1−sin⁡x=limx→0 2cos⁡x×11+sin⁡x+1−sin⁡x=2×1×11+1=1.limx→∞ g(x)=limx→∞ 1+1xx×1x+1x2=e0=1

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.

limx→0 f(x)=limx→0 1+sin⁡x−1+sin⁡xtan⁡x×11+sin⁡x+1−sin⁡x=limx→0 2cos⁡x×11+sin⁡x+1−sin⁡x=2×1×11+1=1.limx→∞ g(x)=limx→∞ 1+1xx×1x+1x2=e0=1

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Let $f (x) = \frac{\sqrt{1 + \sin x} - \sqrt{1 - \sin x}}{\tan x}, x \neq 0$ and $g (x) = {(1 + \frac{1}{x})}^{\frac{x + 1}{x}}$
Statement- $1 :$ $lim_{x \to 0} f (x) = lim_{x \to \infty} g (x)$
Statement- $2 :$ Both the limits are equal to $1 .$

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STATEMENT-1 is True, STATEMENT-2 is False

STATEMENT-1 is False, STATEMENT-2 is Tru

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

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

answer is A.

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Detailed Solution

$lim_{x \to 0} f (x)$

$\begin{array}{l} = lim_{x \to 0} \frac{1 + \sin x - 1 + \sin x}{\tan x} \times \frac{1}{\sqrt{1 + \sin x} + \sqrt{1 - \sin x}} \\ = lim_{x \to 0} 2 \cos x \times \frac{1}{\sqrt{1 + \sin x} + \sqrt{1 - \sin x}} \\ = 2 \times 1 \times \frac{1}{1 + 1} = 1 . \\ \begin{array}{l} lim_{x \to \infty} g (x) = lim_{x \to \infty} {(1 + \frac{1}{x})}^{x \times (\frac{1}{x} + \frac{1}{x^{2}})} = e^{0} = 1 \end{array} \end{array}$