If f(x)=limn→∞(cosxn)n, g(x)=limn→∞(1−x+xen)n & h(x)=tan−1(g−1(f−1(x))), then which of the following is incorrect?

f(x)=limn→∞(cosxn)n=limn→∞e(cosxn−1)n=e−x2 x≥0⇒e−x2≤1 g(x)=limn→∞(1−x+xen)n=elimn→∞x(e1n−1)1n=ex h(x)=tan−1(g−1(f−1(x)))=tan−1(ln(−2lnx)) limx→0+ln(g(x))ln(f(x))=limx→0+x−x2=−2

If f(x)=limn→∞(cosxn)n, g(x)=limn→∞(1−x+xen)n & h(x)=tan−1(g−1(f−1(x))), then which of the following is incorrect?

f(x)=limn→∞(cosxn)n=limn→∞e(cosxn−1)n=e−x2 x≥0⇒e−x2≤1 g(x)=limn→∞(1−x+xen)n=elimn→∞x(e1n−1)1n=ex h(x)=tan−1(g−1(f−1(x)))=tan−1(ln(−2lnx)) limx→0+ln(g(x))ln(f(x))=limx→0+x−x2=−2

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If $f (x) = \lim_{n \to \infty} {(\cos \sqrt{\frac{x}{n}})}^{n},$ $g (x) = \lim_{n \to \infty} {(1 - x + x \sqrt[n]{e})}^{n}$ & $h (x) = \tan^{- 1} (g^{- 1} (f^{- 1} (x)))$ , then which of the following is incorrect?

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$\lim_{x \to 0^{+}} \frac{\ln (g (x))}{\ln (f (x))} = \frac{1}{2}$

$g (x) > 0$

Range of $f (x)$ is $(0, 1]$

Range of $h (x) i s - \frac{π}{2} < h (x) < \frac{π}{2}$

answer is A.

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Detailed Solution

$f (x) = \lim_{n \to \infty} {(\cos \sqrt{\frac{x}{n}})}^{n} = \lim_{n \to \infty} e^{(\cos \sqrt{\frac{x}{n}} - 1) n} = e^{- \frac{x}{2}} x \geq 0 \Rightarrow e^{- \frac{x}{2}} \leq 1 g (x) = \lim_{n \to \infty} {(1 - x + x \sqrt[n]{e})}^{n} = e^{\lim_{n \to \infty} x \frac{(e^{\frac{1}{n}} - 1)}{\frac{1}{n}}} = e^{x} h (x) = \tan^{- 1} (g^{- 1} (f^{- 1} (x))) = \tan^{- 1} (\ln (- 2 \ln x)) \lim_{x \to 0^{+}} \frac{\ln (g (x))}{\ln (f (x))} = \lim_{x \to 0^{+}} \frac{x}{- \frac{x}{2}} = - 2$