What is the limits?

Question

Indeterminate Forms

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Answer 1 · 2017-10-23T18:26:37

$-1/2$ via L'hopital's rule, see 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.

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