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anonymous
 one year ago
Trigonometric limit help.
anonymous
 one year ago
Trigonometric limit help.

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I can't figure out how to fix the denominator in any way.

freckles
 one year ago
Best ResponseYou've already chosen the best response.4\[\lim_{x \rightarrow 0} \frac{x^3 \sin(2x)}{\cos(x^4\frac{\pi}{2}) \tan(x^3+x)}\] writing it here so I don't have to keep looking back and forth and now thinking...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yep no problems haha.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Take a look at my work too. I figured out almost all of it.

freckles
 one year ago
Best ResponseYou've already chosen the best response.4what happens when you plug in 0 we should get 0/0 so we definitely have more work to do...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0It's just that pesky first limit.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0If I was allowed to use taylor polynomials this would be a joke but I am not.

freckles
 one year ago
Best ResponseYou've already chosen the best response.4yeah I was looking at your work and that one term we have a problem with since that first limit doesn't actually exist \[\lim_{x \rightarrow 0} \frac{x^3\cos(x^3+x)(2x)}{x^4(x^3+x)} \\ \lim_{x \rightarrow 0} 2 \frac{\cos(x^3+x)}{x^3+x} \\ \text{ plug \in 0 you get } \frac{1}{0} \\ \text{ so the limit does not exist } \\ \text{ so I don't think we can take the limit of each factor }\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.4you get 2/0 sorry forgot to multiply the 2 on top to the 1 on top

freckles
 one year ago
Best ResponseYou've already chosen the best response.4still doesn't exist though

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0According to wolfram, it's 4 though.

freckles
 one year ago
Best ResponseYou've already chosen the best response.4for the one I just said doesn't exist?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0OMG> Sorry correction. The sin(2x) in the numerator should be a sin^2(2x) .

freckles
 one year ago
Best ResponseYou've already chosen the best response.4so the work on that one page is slightly off then because of this new factor sin(2x)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yep let me try again...

freckles
 one year ago
Best ResponseYou've already chosen the best response.4\[\lim_{x \rightarrow 0} \frac{ x^3 \cos(x^3+x) (2x) \sin(2x)}{x^4(x^3+x)} \\ \lim_{x \rightarrow 0} \frac{\sin(2x)}{2x} (2) \frac{x^3 \cos(x^3+x)(2x)}{x^3(x^3+x) } \\ \lim_{x \rightarrow 0} 2 \frac{\cos(x^3+x)}{x^3+x}(2x) \\ \lim_{x \rightarrow 0 } 4 \frac{\cos(x^3+x)}{x^2+1}\] plug in 0 :)

freckles
 one year ago
Best ResponseYou've already chosen the best response.4and dido awesome algebra going on your work

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Omg thanks T_T . I can't believe a typo threw me off.

freckles
 one year ago
Best ResponseYou've already chosen the best response.4yep throwing in that sin(2x) into your first factor there totally fixed it so the limit would exist
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