Question

I:$\underset{n\to \infty }{lim}\frac{n!}{(n+1)!-n!}=0$
II:$\underset{x\to 0^{+}}{lim}\sqrt{\frac{x-{\sin }^{2}x}{x+\cos x}}=1$

• 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

The correct option is A Only I is true

I.$\underset{n\to \infty }{lim}\frac{n!}{(n+1)!-n!}=\underset{n\to \infty }{lim}\frac{n!}{n!(n+1)-n!}=\underset{n\to \infty }{lim}\frac{1}{n}=0$
II. $\underset{x\to 0^{+}}{lim}\sqrt{\frac{x-{\sin }^{2}x}{x+\cos x}}=\underset{x\to 0^{+}}{lim}\sqrt{\frac{0-{\sin }^{2}0}{0+\cos 0}}$ (using direct substitution)
$=0$