Question

Use L'hopitals rule to find the limit. Limit as x approaches infinity. x sin (19/x)

Answer 1

So to use l'Hopital's Rule, you will need to write your function as a ratio of two functions both of which have limit of 0 as x --> infty.

Can you see how to write it that way?

Answer 2

LIM X
LIM SIN 19/X?

Answer 3

\[ \lim_{x \to \infty} x = \infty \] not zero

Answer 4

ok.

Answer 5

so then the same with lim 19/infinity.

Answer 6

To reiterate, we need to write
\[ x \sin(19/x) = \frac{f(x)}{g(x)} \]

where
\[ \lim_{x \to \infty} f(x) = \lim_{x \to \infty} g(x) = 0 \]

Answer 7

Then by l'Hopital's rule, provided the appropriate limits exist,
\[ \lim_{x \to \infty} \frac{f(x)}{g(x)} = \lim_{x \to \infty} \frac{f'(x)}{g'(x)} \]

Answer 8

So what f(x) and g(x) would do this?

Answer 9

what would make g(x) and f(x) equal 0?

Answer 10

not equal to zero, having limits equal to zero.

Answer 11

hint: f(x) = sin(19/x). Now, what's g(x) ?

Answer 12

If f(x) = sin(19/x) and f(x)/g(x) = x.sin(19/x), solve for g(x).

Answer 13

1/sin (19/x)

Answer 14

\[ \frac{f(x)}{g(x)} = \frac{\sin(19/x)}{g(x)} = x \sin(19/x) \]
Therefore \( g(x) = 1/x \).

Now, is it the case that the limits of f(x) and g(x) as x --> infty are both zero?

Answer 15

they both get closer to 0

Answer 16

they get smailler and smaller

Answer 17

their limits are both zero. Now apply l'Hopital's rule. Calculate
\[ \lim_{x \to \infty} \frac{f'(x)}{g'(x)} \]

Answer 18

so the answer is 0

Answer 19

No.

Answer 20

f'(x) is -19 cos (19/x) / x^2

Answer 21

Yes and g'(x) = ...?

Answer 22

and the g'(x) is 1/x^2

Answer 23

No, -1/x^2

Answer 24

oops that what i meant.

Answer 25

Hence f'(x)/g'(x) = ...?

Answer 26

19 cos (19/x)

Answer 27

Yes and therefore
\[ \lim_{x \to \infty} \frac{f'(x)}{g'(x)} = ...? \]

Answer 28

19

Answer 29

Yes.

Answer 30

sweet thanks sorry it took me a while. I really appreciate it!