Question

What is the limit of `sinx /(x^2 - 4x)` as x approaches 0?

Answer 1

`lim_(x->0)sinx/(x^2-4x) = -1/4`

Explanation:

You know that `sin0 = 0`, so you cannot evaluate this limit by replacing `x` with zero, since that would get you

`color(red)(cancel(color(black)(lim_(x->0)(sin0)/(0^2 - 4 * 0) = 0/0))) ->` indeterminate form

This means that you're going to have to use L'Hopital's Rule to find this limit. According to L'Hopital's rule, if you have two functions `f(x)` and `g(x)` which are differentiable on an interval `(a,b)`, and if

`lim_(x->c)(f^'(x))/(g^'(x))" "`, with `g^'(x)!=0` and `c in (a,b)`

exists, then you have

`color(blue)(lim_(x->0)(f(x))/(g(x)) = lim_(x->0)(f^'(x))/(g^'(x))`

In your case, you have

`d/dx(sinx) = cosx" "` and `" "d/dx(x^2-4x) = 2x-4`

which means that

`lim_(x->0)sinx/(x^2-4x) = lim_(x->0)(d/dx(sinx))/(d/dx(x^2-4x)) = cosx/(2x-4)`

Now you can evaluate this limit for `x->0` by replacing `x` with zero

`lim_(x->0)(cos0)/(2 * 0 - 4) = 1/((-4)) = -1/4`

Therefore

`lim_(x->0)sinx/(x^2-4x) = color(green)(-1/4)`

Answer 2

`lim_(xrarr0) sinx/(x^2-4x) = -1/4`

Explanation:

`lim_(xrarr0) sinx/(x^2-4x) = lim_(xrarr0) sinx/(x(x-4))`

`= lim_(xrarr0) sinx/x 1/((x-4))`

`= lim_(xrarr0) sinx/x lim_(xrarr0) 1/((x-4))` `" "`if both limits exist

`= (1)(-1/4)`

`= -1/4`