# Thread: A simple question based on limits

1. ## A simple question based on limits

i know the theorem that,
when lim x tends to 0, then (sinx)/x=1.
then my question is

when lim x tends to 0, then what is sinx ?

is the answer 0 or x?

(i just know that we can multiply and divide by x. then (sinx)/x is 1 by theorem, then what will happen to the x which is multiplied in the numerator ? Limit will be also applied to that and the answer of the question would be 0. isn't it ? is it right or wrong ? )

2. Originally Posted by iamtheone
i know the theorem that,
when lim x tends to 0, then (sinx)/x=1.
then my question is

when lim x tends to 0, then what is sinx ?

is the answer 0 or x?

(i just know that we can multiply and divide by x. then (sinx)/x is 1 by theorem, then what will happen to the x which is multiplied in the numerator ? Limit will be also applied to that and the answer of the question would be 0. isn't it ? is it right or wrong ? )

Let's look at the region of $x \in [0, \frac{\pi}{2}]$

$sin(x) \le x \le tan(x)$

$\frac{sin(x)}{sin(x)} \le \frac{x}{sin(x)} \le \frac{tan(x)}{sin(x)}$

$1 \le \frac{x}{sin(x)} \le \frac{sin(x)}{sin(x)cos(x)}$

$\frac{x}{sin(x)} \le \frac{1}{cos(x)}$

$lim_{x \rightarrow 0} \frac{1}{cos(x)} = 1$

Using the squeeze theorem:

$lim_{x \rightarrow 0} \frac{x}{sin(x)} = 1$