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

A
STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is a correct explanation for STATEMENT-1
B
STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is NOT a correct explanation for
STATEMENT-1
C
STATEMENT-1 is True, STATEMENT-2 is False
D
STATEMENT-1 is False, STATEMENT-2 is True

Solution:

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

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

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

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

Solution:

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