Question

Which of the following limits is/are unity?

• A. $\underset{x\to \pi }{lim}\frac{\sin (\pi -x)}{(\pi -x)}$
• B. $\underset{x\to 0}{lim}\frac{\tan \left(\sin x\right)}{\tan x}$
• C. $\underset{x\to \infty }{lim}\frac{x^2+x}{x^2-x}$
• D. $\underset{x\to 1}{lim}\frac{\tan \left(\pi x\right)}{\pi (x-1)}$

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A $\underset{x\to 0}{lim}\frac{\tan \left(\sin x\right)}{\tan x}$
B $\underset{x\to \infty }{lim}\frac{x^2+x}{x^2-x}$
C $\underset{x\to 1}{lim}\frac{\tan \left(\pi x\right)}{\pi (x-1)}$
D $\underset{x\to \pi }{lim}\frac{\sin (\pi -x)}{(\pi -x)}$

A) $\underset{x\to \pi }{lim}\frac{\sin (\pi -x)}{(\pi -x)}=1$
B) $\underset{x\to 0}{lim}\frac{\tan \left(\sin x\right)}{\tan x}=\underset{x\to 0}{lim}(\frac{\sin \left(\sin x\right)}{\sin x}\times \frac{\cos x}{\cos \left(\sin x\right)})$

$=\underset{x\to 0}{lim}\frac{\sin \left(\sin x\right)}{\sin x}\times \underset{x\to 0}{lim}\frac{\cos x}{\cos \left(\sin x\right)}=1\times 1=1$
C) $\underset{x\to \infty }{lim}\frac{x^2+x}{x^2-x}=\underset{x\to \infty }{lim}\frac{1+\frac{1}{x}}{1-\frac{1}{x}}=1$
D) $\underset{x\to 1}{lim}\frac{\tan \left(\pi x\right)}{\pi (x-1)}=\underset{x\to 1}{lim}\frac{\sin \left(\pi x\right)}{\pi \cos \left(\pi x\right)(x-1)}$

$=\frac{1}{\pi }\underset{x\to 1}{lim}\frac{1}{\cos \left(\pi x\right)}\times \underset{x\to 1}{lim}\frac{\sin \left(\pi x\right)}{(x-1)}$

$=-\frac{1}{\pi }\underset{x\to 1}{lim}\frac{\sin \left(\pi x\right)}{(x-1)}=ApplyLHospitalRule\Rightarrow -\underset{x\to 1}{lim}\frac{\pi \cos \left(\pi x\right)}{\pi }=1$