Applying L'hôpitals gets you lim x->0 (e^x-1)/1, which you can then plug x->0 into to get (e^0-1)/1 = (1 - 1)/1 = 0. To do this without L'hôpitals, you might find another sequence that approaches zero and prove that each term in the first sequence is closer to zero than the corresponding term in the second, but it's too late for me to do that right now.
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page 7 jojo reference
Other limit Quiz for mathematicianz
lim x->0 (e^x-1)/x = ?
(Without L'hôpitals!!!)
Applying L'hôpitals gets you lim x->0 (e^x-1)/1, which you can then plug x->0 into to get (e^0-1)/1 = (1 - 1)/1 = 0. To do this without L'hôpitals, you might find another sequence that approaches zero and prove that each term in the first sequence is closer to zero than the corresponding term in the second, but it's too late for me to do that right now.
Hold up, the 'x-' being directly before the '>' makes this equation illogical.
I think it means x goes to zero as limit
Correct, this his how limes works.
Infinite
Looking forward to more!
I just can't jack it to this the girl looks to much like Ace the ocarina maker's starbound character
Is page 10 a jojo reference??
She knows the secret joestar technique
These lines are so awkward
nice it do not get my seal butt ok
The x in page 17 is 5 hahaha
WHY DO THINGS LIKE THIS KEEP POPPING UP IN PORN COMMENT SECTIONS!!!!?????
You're not solving for x, you're finding the limit of the function as x approaches -3, which I think is -6/10
Yeah, thats right
Finally someone smart
6+4x / x^2 + 1 =(divide by x both sides)= 6+4 / x + 1 =(simplify)= 10 / x + 1 =(get x on its own)= 9 / x === answer is 9 over x
None of that is correct.
Your thinking of algebra. -0.6 is the answer. But good try though
To short but good
This is good
It's - 0.6 right? (haven't seen limes to a number yet, only with infinity, so I'm just guessing)
Yeah it is.
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