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anonymous
 4 years ago
lim x^2cos2x/1cosx
x>0
Can someone explain to me how you solve this?
anonymous
 4 years ago
lim x^2cos2x/1cosx x>0 Can someone explain to me how you solve this?

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slaaibak
 4 years ago
Best ResponseYou've already chosen the best response.2Use L'Hospitals rule. If the limit is in an indeterminate form, the limit of f(x)/g(x) is equal to f'(x)/g'(x)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I dont think I cant use that they'll mark me off on my test. We can only use calc I stuff.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0im suppose to use something along the lines of lim(h>0) sinh/h=1 or lim(h>0) cosh1/h=0, I'm guessing because thats what they put in the examples.

slaaibak
 4 years ago
Best ResponseYou've already chosen the best response.2You should be able to do partition it like that. I'll do it quickly

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0the only rule i can see that applies is the quotient rule lim(x>c) f(x)/g(x)= lim f(x)/ lim g(x) if g(x) not equal to zero

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.1\[\frac{x^2 \cos(2x)}{1\cos(x)}=\frac{x^2 \cos(2x)}{1\cos(x)} \frac{1+cos(x)}{1+cos(x)}\] \[\frac{x^2 \cos(2x) (1+\cos(x))}{1\cos^2(x)}\] \[\frac{x^2 \cos(2x)}{\sin^2(x)}=\frac{x^2}{\sin^2(x)} \cdot \cos(2x) =(\frac{x}{\sin(x)})^2 \cdot \cos(2x)\]

slaaibak
 4 years ago
Best ResponseYou've already chosen the best response.2\[\lim_{x \rightarrow 0} {x^2 \cos2x \over 1 \cos x} = \lim_{x \rightarrow 0} {x^2 \cos 2x (1+ cosx) \over \sin^2 x}\] Meh too late. What myinanaya said!

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.1oops i left over my (1+cos(x))

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.1sorry i'm on my slow laptop

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0myininaya is the master of these trig limits

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0sin^2x when x=0 is 0 that would make it undefined though

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0or even with sinx on the bottom

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.1\[\lim_{x \rightarrow 0}\frac{x}{\sin(x)}=1\]

slaaibak
 4 years ago
Best ResponseYou've already chosen the best response.2I didn't complete mine as myin was before me

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.1I just left off a factor by the way (1+cos(x)) so don't forget to look at that when you are doing this @ awstin

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0makes you appreciate l'hopital

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I get it now thank you, I didnt know that x/sinx = 1 though lol

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0l'hopital makes this easier?

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.1you actually prove that on the way to proving sin(x)/x>1 as x>0 when using squeeze thm

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0cacl is hard :\ I was amazing at precalc too!
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