Question

$\underset{x\to 0}{lim}\frac{sinx}{x}=1$ Prove the given limit using sandwich theorem. To see solution press yes.

• A. yes
• B. No

Answer 1

The correct option is B

No

Statement of sandwich theorem:

If f,g and h be real function such taht $f\left(x\right)\le g\left(x\right)\le h\left(x\right)$ for all x in the common domain.

if $\underset{x\to a}{lim}f\left(x\right)=l=\underset{x\to a}{lim}h\left(x\right)$ then $\underset{x\to a}{lim}g\left(x\right)=l$

Lets take three function

$f\left(x\right)=cosx$

$g\left(x\right)=sinx$

$h\left(x\right)=1$

Since, sin x & cos x both are positive in the inteval $(0,\frac{\pi }{2})$

Lets's take the interval on the domain $(0,\frac{\pi }{2})$

O is the centre of unit circle such that $\angle AOC$ is x radian and 0 < x < $\frac{\pi }{2}$