Question

Question #d5a95

Answer 1

This question is answered here: https://socratic.org/questions/how-do-you-use-the-squeeze-theorem-to-find-lim-sin-x-x-as-x-approaches-zero

Answer 2

See the explanation.

Explanation:

We must prove that

`color(brown)(lim_(x to 0)sin(x)/x =1`

L'Hospital's Rule is a handy tool for calculating limits involving indeterminate forms like

`color(green)(0/0 or (+- oo)/(+-oo)`

If we are taking the limit

`color(blue)(lim_(x to a)f(x)/g(x),` we get `color(blue)(f(a)/g(a) = 0/0 or (+- oo)/(+-oo)`

Apparently, when we use substitution method to evaluate the limit, we end up with an indeterminate form

L'Hospital's Rule states that instead of evaluating

`color(blue)(lim_(x to a)f(x)/g(x)`

we can evaluate the limit of

`color(blue)(lim_(x to a)(f'(x))/(g'(x)`

Given:

`lim_(x to 0) [sin(x)/x]`

When we evaluate this limit we get `sin(0)/0 = 0/0`, an indeterminate form.

Hence, we will use the L'Hospital's Rule for our problem.

`lim_(x to 0) [[sin(x)']/[(x)']]`

Differentiate `sin(x)`

`(dy)/(dx) sin(x) = cos(x)`

and

`(dy)/(dx) (x) = 1`

We will use these intermediate results in

`lim_(x to 0) [[sin(x)']/[(x)']]`

We get,

`lim_(x to 0) [[cos(x)]/[1]]`

`rArr lim_(x to 0) cos(x)`

Substitute `x = 0` to get

`Cos(0) = 1`

Hence,

`color(brown)(lim_(x to 0)sin(x)/x =1`