anonymous
  • anonymous
limit as x approaches 0 of ln(secx)x^4
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
Directly substituting should do the trick
anonymous
  • anonymous
I got 0, how about if x is approaching infinity, would that be the indeterminate form infinity times 0?
anonymous
  • anonymous
Yes, in that case you have to rewrite the function so that you get a proper indeterminate form.

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anonymous
  • anonymous
For this particular function, though, \(\displaystyle\lim_{x\to\infty}\sec x\) does not exist, so there is no indeterminate form involved here.
anonymous
  • anonymous
I don't get how it doesn't exist?
anonymous
  • anonymous
Think of it this way: \[\lim_{x\to\infty}x^4\ln\sec x=\lim_{x\to\infty}x^4\cdot\lim_{x\to\infty}\ln\sec x\] Using the continuity of the natural log, you have \[\lim_{x\to\infty}x^4\cdot\ln\left(\lim_{x\to\infty}\sec x\right)\] The first limit is obviously ∞, but the second does not exist.
anonymous
  • anonymous
|dw:1377738320412:dw| As x grows larger and larger, there are multiple asymptotes.
anonymous
  • anonymous
I see what you mean thanks
anonymous
  • anonymous
you're welcome
anonymous
  • anonymous
for the limit as x approaches infinity of x^2 times e^(-2x), its indeterminate form 0 times -infinity I move the x^2 to the bottom right. Then I simplify and use L'Hopitals rule right?
anonymous
  • anonymous
You could, but it'd be easier to move the e^(-2x). \[e^{-2x}=\frac{1}{e^{2x}}\]
anonymous
  • anonymous
\[\lim_{x\to\infty}x^2e^{-2x}=\lim_{x\to\infty}\frac{x^2}{e^{2x}}=\frac{\infty}{\infty}\] Apply L'hopital's rule as many times as necessary.
anonymous
  • anonymous
Thank you, the infinity sign confuses me a lot

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