QUICK EASY QUESTION

l'hopital's rule lol

>>9067942
Is that Hairtony Linetano?

>>9067947
Without using l'hopital.

>>9067947
For example in similar problems i'd multiply using the conjugate of the numerator. Until i could plug x in and get something other then zero.

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Kys.

use: (true for any functions ψ)
lim ψ->0 ( sin ψ / ψ ) = 1, and:
tan ψ = sin ψ / cos ψ , so:

lim x->0 (sin 2x tan 3x / x^2) =
lim x->0 2*3(sin 2x / 2x)(sin 3x / 3x)(1/cos3x) =
lim x->0 6*(1/cos 3x) = 6

>>9068015
Thanks senpai.

>>9068016
Now i'm confused about how you went from

"(sin 2x tan 3x / x^2)" to "2*3(sin 2x / 2x)(sin 3x / 3x)(1/cos3x)"

>>9068015
This is wrong lul

>>9068058
He did a few steps, what is happening there is reareanging it into the form of sin theta / theta so that it goes to one.

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>>9068015
Sheeeeeit

>>9068133
Nope, it's valid. He just implicitly used sinx/x = tanx/x = 1 as x approaches 0.