Question

evaluate limit:
lim x approaches 0 e^(5x)-1-5x/x^2....

Answer 1

no i got it to be:
5(e^5x)-1-5x/2x

Answer 2

Consider only x/x^2
Because the bottom exponent greater, it goes to 0

Answer 3

i got the answer to be 5/2....

Answer 4

e^5x=1
So 1-1=0. The lim is 0

Answer 5

yes...and..

Answer 6

and..what?

Answer 7

i said the limit is 0...but i wanna know what the answer is so that i can follow it up...i dont know wher 5/2 come from

Answer 8

Oh, sorry I was reading the rules for infinity. 0/0 Lhopital

Answer 9

Lhopital produces 5/x
That lim goes to infinity.

Answer 10

yes i also derivtive x^2 to be 2x...

Answer 11

5/2x
lim is infinity

Answer 12

ok thanks : )

Answer 13

what?

Answer 14

\[\lim_{x \rightarrow 0} \frac{ e^{5x} -1 -5x}{x^2} \neq \infty\]

Answer 15

differentiate top and bottom once

Answer 16

\[\frac{5e^{5x} -5}{2x} \]

Answer 17

still 0/0

differentiate again

Answer 18

\[\frac{25e^{5x}}{2}\]

Answer 19

now sub x=0

Answer 20

\[\frac{25}{2}\] is the answer

Answer 21

ok thanks

Answer 22

I thought only the 5x was over x^2. Use brackets rmalik

Answer 23

Firstlt, after pluging 0, we get 0/0, there fore, we need to take the derivative of both the denomenator and nominator, then, we will find that it still 0, take the second derivative and you will get 25/2

e^5x-1-5x/x^2 ==>5e^5x-5/2x ==>25e^5x/2

now, plug 0 into the equation and you will get 25/2 !