Ill go one at a time here
lim x -> 0
(e^(2x) - 1) / e^(x)-1
yeah I have no idea how to do this without stuff I shouldn't know
Ill go one at a time here
lim x -> 0
(e^(2x) - 1) / e^(x)-1
yeah I have no idea how to do this without stuff I shouldn't know
-1
2
it's a simple thing when you learn it, l'hopital's rule. if you have a limit where when you plug in the value its 0/0 or infinity/infinity, take the individual derivatives of the top and bottom until you get something not undefined
69
Im not suppose to know that yet God. If I do that on a test that isn't the right work for the answer. Its like those kids in elementary school that could multiply large numbers in their head, but still needed to write down the organized process.
I thought you meant (e^(2x) - 1) / e^(x)) * -1 for some reason
use the squeeze theorem. take something like 10000000000x^2 + 2 and -100000000000x^2 + 2. they're going to be strictly above and below your function respective on, say (-1, -1), and both their limits at 0 are 2. this your function also tends to 2 at 0 since its continuous.
i mean, that's a proof moreso a method to find the answer. but you could always just find the answer for yourself with l'hopital and make yourself appropriate functions to justify your answer with the squeeze theorem on the test
God your avatar reminds me a lot of my math teacher:
Kirby the Racist
Yeah I can never spot the difference between the Japanese and Chinese races.
The Japanese were business suits, and the Chinese have pony tails
Is it just me and my asian fetish, or does the one is the middle look very well drawn?