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
"I'll go," said Chagataev. "But what will I do there? Build socialism?"
"What else?" said the secretary.
69
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'll go," said Chagataev. "But what will I do there? Build socialism?"
"What else?" said the secretary.
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
"I'll go," said Chagataev. "But what will I do there? Build socialism?"
"What else?" said the secretary.
"I'll go," said Chagataev. "But what will I do there? Build socialism?"
"What else?" said the secretary.
Kirby the Racist