anonymous
  • anonymous
Trigonometric limit help.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
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anonymous
  • anonymous
I can't figure out how to fix the denominator in any way.
freckles
  • freckles
\[\lim_{x \rightarrow 0} \frac{x^3 \sin(2x)}{\cos(x^4-\frac{\pi}{2}) \tan(x^3+x)}\] writing it here so I don't have to keep looking back and forth and now thinking...

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anonymous
  • anonymous
Yep no problems haha.
anonymous
  • anonymous
Take a look at my work too. I figured out almost all of it.
freckles
  • freckles
what happens when you plug in 0 we should get 0/0 so we definitely have more work to do...
anonymous
  • anonymous
It's just that pesky first limit.
anonymous
  • anonymous
If I was allowed to use taylor polynomials this would be a joke but I am not.
freckles
  • freckles
yeah 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
  • freckles
you get 2/0 sorry forgot to multiply the 2 on top to the 1 on top
freckles
  • freckles
still doesn't exist though
anonymous
  • anonymous
According to wolfram, it's 4 though.
freckles
  • freckles
for the one I just said doesn't exist?
anonymous
  • anonymous
WAIT.
anonymous
  • anonymous
OMG> Sorry correction. The sin(2x) in the numerator should be a sin^2(2x) .
freckles
  • freckles
so the work on that one page is slightly off then because of this new factor sin(2x)
anonymous
  • anonymous
Yep let me try again...
freckles
  • freckles
\[\lim_{x \rightarrow 0} \frac{ x^3 \cos(x^3+x) (2x) \sin(2x)}{x^4(x^3+x)} \\ \lim_{x \rightarrow 0} \frac{\sin(2x)}{2x} (2) \frac{x^3 \cos(x^3+x)(2x)}{x^3(x^3+x) } \\ \lim_{x \rightarrow 0} 2 \frac{\cos(x^3+x)}{x^3+x}(2x) \\ \lim_{x \rightarrow 0 } 4 \frac{\cos(x^3+x)}{x^2+1}\] plug in 0 :)
freckles
  • freckles
and dido awesome algebra going on your work
anonymous
  • anonymous
Omg thanks T_T . I can't believe a typo threw me off.
anonymous
  • anonymous
Haha thanks!
freckles
  • freckles
yep throwing in that sin(2x) into your first factor there totally fixed it so the limit would exist

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