Question

What is the limits?

Indeterminate Forms

Answer 1

`-1/2` via L'hopital's rule, see explanation

Explanation:

We will use L'hopital's rule, which states that:

`lim_(x->a) f(x)/g(x) = lim_(x->a) (f'(x))/(g'(x))`.

So for `f(x) = 1-e^(2x), f'(x) = -2e^(2x), and with g(x) = sin5x - x, g'(x) = 5cos5x - 1`, giving us...

`lim_(x->0) (-2e^(2x))/(5cos5x - 1)`

Now we can plug in `x=0`, and see what we get. If we're still in indeterminate form, we may have to use a second iteration of the rule.

`= (-2e^0)/(5cos 0 -1) = -2/(5-1) = -2/4 = -1/2`