Question

$I:\underset{x\to 0}{lim}(\frac{\tan \left[e^2\right]x^2-\tan [-e^2]x^2}{{\sin }^{2}x})=15$
$II:\underset{x\to 0}{lim}(\frac{1^x+2^x+3^x+\dots \dots +n^x}{n}{)}^{\frac{1}{x}}=(n!{)}^{1/n}$

• A. Only I is true
• B. Only II is true
• C. Both I and II are true
• D. Neither I nor II is true

Answer 1

Answer: (C)

Both I and II are true

$I.\underset{x\to 0}{lim}\left(\frac{\tan \left[e^2\right]x^2-\tan \left[{-e}^2\right]x^2}{{\left(\sin x\right)}^2}\right)$

$=\underset{x\to 0}{lim}(\frac{\tan \left[e^2\right]x^2}{x^2}-\frac{\tan \left[{-e}^2\right]x^2}{x^2})\times \frac{x^2}{{\left(\sin x\right)}^2}$

$=(\left[e^2\right]-[-e^2])\times 1\left[\underset{x\to 0}{lim}\frac{\tan ax}{x}=a\right]$

$=7-(-8)[e\approx 2.71,e^2\approx 7.3]$

$=15$

$II.\underset{x\to 0}{lim}{\left(\frac{1^x+2^x+3^x+....+n^x}{n}\right)}^{\frac{1}{x}}-1^{\infty }$form

$=e^{\underset{x\to 0}{lim}\frac{1}{x}(\frac{1^x+2^x+3^x+....+n^x}{n}-1)}$

$=e^{\underset{x\to 0}{lim}\frac{1}{x}(\frac{1^x-1}{n}+\frac{2^x-1}{n}+...+\frac{n^x-1}{n})}$

$=e^{\frac{1}{n}\underset{x\to 0}{lim}(\frac{1^x-1}{n}+\frac{2^x-1}{n}+...+\frac{n^x-1}{n})}$

$=e^{\frac{1}{n}(\log 1+\log 2+....+\log n)}$

$=e^{\frac{1}{n}\left(\log \mathrm{1.2.3....}n\right)}=e^{\log {(n!)}^{1/n}}$

$={(n!)}^{1/n}$