L- Hospitals Rule CALC 1?

Question

This is one more i am stuck on. L- Hospitals Rule

Find the limlt. Use L'Hospital's Rule where appropriate. If there is a more elementary method, consider using it. If L'Hospital's Rule doesn't apply, expla!n why.

1. lim t-->0 for ( e^(3t) - 1 ) / t

Answer 1

Ok lim e^(3t) - 1 as t ----> 0 = e^(0) - 1 = 1 - 1 = 0

lim (e^(3t) - 1) / t as t -----> 0 = 0 / 0 which is undefined, L'hospital rule is applicable here (Remember that this rule is only applicable when you have a 0 /0 format or infinity / infinity).

The limit becomes:
lim 3e^(3t) / 1 as t --->0 = 3e^0 = 3

So lim (e^(3t) - 1) / t as t ----> 0 = 3

Answer 2

L'Hospital's rule says that given a function h(x) = f(x)/g(x) where f(x) -> 0 and g(x) -> 0 when x-> 0, you can find the limit by evaluating f'(x)/g'(x) as x->0 where the ' menas derivative with respect to x.

SO let's use this rule:

lim t->0 (e^(3t)-1)/t as t-> 0 e^(3t)-1 ->0 and t->0 so we can use the rule.

d(e^(3t)-1)/dt = 3e^(3t) and d(t)/dt = 1

Thus:
lim t->0 (e^(3t)-1)/t = lim t -> 0 3e^(3t)/1 = 3

Answer 3

First, plug in 0 for t and see what you get.

( e^(3*0) - 1) / 0 = (1-1) / 0 = 0/0

We have 0/0 indeterminate form, so use L'Hopital's Rule and differentiate the top and the bottom.

d/dx (e^(3t)-1) / d/dx (t) = 3e^(3t) / 1 = 3e^(3t)

Now plug in 0 again.

3e^(3*0) = 3

Answer 4

Lim = d/dx [e^(3t)-1] / d/dx (t)
t --> 0

use the chain rule:
let u = 3t
then d/du e^u = e^u
d/dx (3t) = 3

lim = 3e^(3t)
t --> 0

lim = 3e^(3*0)
t --> 0

lim = 3 <== answer
t --> 0