Question

whose solve first ?​

Answer 1

Answer:

$\large\boxed{\sf{1}}$

Step-by-step explanation:

Given a limit such that,

$\displaystyle \lim_{x\to0} {( \frac{ \sin x }{x} )}^{ \frac{1}{ {x}^{2} } }$

When we will put the limit, we come to find that it's a form of $\bold{{1}^{\infty}}$

Now, this type of limits are solved by this method.

Therefore, we will get,

$= {e}^{\displaystyle \lim_{x\to0} \frac{1}{ {x}^{2} } ( \frac{ \sin x}{x} - 1)}$

Now, let's solve the limit which is in power to the exponential.

Therefore, we will get,

$= \displaystyle \lim_{x\to0} \frac{1}{ {x}^{2} } ( \frac{ \sin x}{x} - 1) \\ \\ = \frac{1}{ {x}^{2} } (\displaystyle \lim_{x\to0} \frac{ \sin x }{x} - 1)$

But, we know that,

• $\displaystyle \lim_{x\to0} \frac{ \sin(x) }{x} = 1$

Therefore, we will get,

$= (1 - 1)\displaystyle \lim_{x\to0} \frac{1}{ {x}^{2} } \\ \\ = 0 \times ....... \\ \\ = 0$

Thus, the value of power to the exponential is 0.

Therefore, we will get, the value of limit,

$= {e}^{0} \\ \\ = 1$

Hence, the required value of given limit is 1.