Question

What is `lim_(x->-2) sin(5x+10)/(4x+8)` ?

Answer 1

#5/4#

Explanation:

If we plug in `x =-2` we would get `0/0` . This is called "indeterminate form."

The phrase “indeterminate form” is used to mean a function that we can't compute the limit of by simply applying some general theorem.

For instance apply L'Hôpital's theorem , which is stated as:

Suppose that we have one of the following cases,
`lim_{xrarra}f(x)/g(x) = 0/0` or `lim_{xrarra}f(x)/g(x) = {+-oo}/{+-oo}`
where a can be any real number, infinity or negative infinity. In these cases we have,

`lim_{xrarra}f(x)/g(x) = lim_{xrarra}f^'(x)/g^'(x)`

So, L'Hôpital's theorem tells us that if we have an indeterminate form `0/0` or all we need to do is differentiate the numerator and differentiate the denominator and then take the limit.

So by taking derivative of the numerator and the denominator, we will have the limit as:

`=lim_{xrarr-2}((cos(5x+10)*5)/(4))`

Plug in the value `x =-2`

`=(cos(5(-2)+10)*5)/(4)`

Simplify:

`=5/4`

Answer 2

`5/4`

Explanation:

Use:

`lim_(t->0) (sin t)/t = 1`

Putting `t = 5x+10`, we have:

`lim_(x->-2) sin(5x+10)/(4x+8) = lim_(t->0) sin(t)/(4/5t) = 5/4 lim_(t->0) sin(t)/t = 5/4`