Question

Find the limit using lo hospitals rule

Answer 1

..?

Answer 2

limit x->0 (1+4x)^(#/x)

Answer 3

limit x->0 (1+4x)^(3/x)

Answer 4

1

Answer 5

is 1 the anser?

Answer 6

yes

Answer 7

limit of the (3/x) by using l'hospital is 0 and from there limit x-> 1^1=1

Answer 8

sorry 1^0=1

Answer 9

but you should have something diving something to use lopetals

Answer 10

you take derivative of 3 and x

Answer 11

0/1 makes 0

Answer 12

but you don't apply it for 1+4x

Answer 13

Exactly,
take log(to base e) to get
3log(1+4x)/x
and get 12 with l'hôpital's rule.
The limit is therefore e^12.

Answer 14

actually the problem here is for 3/x though

Answer 15

mathmate that is the answer but how did you make it like that
?

Answer 16

i got it

Answer 17

Use l'hôpital's rule on
3log(1+4x) / x
the numerator becomes (3/(1+4x))*4 and the denominator is 1.
When x->0, then the expression becomes 12.
Since we took log(to base e) before, we only have to take the "anti-log" to get e^12.