Question

use l'hopital's rule to find the limit x gos to inf xsin(19/x)

Answer 1

Maybe read the question more carefully. \[\lim_{x \rightarrow \infty} x \sin (19/x)\]doesn't look like a case where you'd use L'Hopital's rule.

Answer 2

sin / (1/x) perhaps

Answer 3

can you explain

Answer 4

or ... x / (csc(19/x))

Answer 5

you are trying to resturcture the setup to a 0/0 form

Answer 6

@amistre64 I like your first one better

Answer 7

the first one might be good since we got a 19/x :) just wasnt all that sure

Answer 8

were did the 19 go in the first one

Answer 9

This is what she meant:
\[\lim_{x \rightarrow \infty} \sin(19/x)/(1/x)\]
Take the derivative of the top and the derivative of the bottom.
Some (1/x^2)s cancel.
limit as cos(19/x) goes to zero is one
The answer should be 19.

Answer 10

\[x~sin(kx)\]
\[\frac{sin(kx)}{1/x}\]
\[\frac{\cancel{cos(k/x)}^1*~\bcancel{-}k\cancel{/x^2}}{\cancel{-1/x^2}}=\]

Answer 11

@amistre64 :)

Answer 12

thanks