Question

limit as x approaches 0 of ln(secx)x^4

Answer 1

Directly substituting should do the trick

Answer 2

I got 0, how about if x is approaching infinity, would that be the indeterminate form infinity times 0?

Answer 3

Yes, in that case you have to rewrite the function so that you get a proper indeterminate form.

Answer 4

For this particular function, though, \(\displaystyle\lim_{x\to\infty}\sec x\) does not exist, so there is no indeterminate form involved here.

Answer 5

I don't get how it doesn't exist?

Answer 6

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.

Answer 7

As x grows larger and larger, there are multiple asymptotes.

Answer 8

I see what you mean thanks

Answer 9

you're welcome

Answer 10

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?

Answer 11

You could, but it'd be easier to move the e^(-2x).
\[e^{-2x}=\frac{1}{e^{2x}}\]

Answer 12

\[\lim_{x\to\infty}x^2e^{-2x}=\lim_{x\to\infty}\frac{x^2}{e^{2x}}=\frac{\infty}{\infty}\]
Apply L'hopital's rule as many times as necessary.