Question

If x->infinity, can L'hopital's rule be applied:
sin(x)/(1+sqrt(x)) ?

Answer 1

It's infinity over infinity right now?

Answer 2

yes

Answer 3

limits are not my strong point i'm afraid
but l'hopitals seems the wat to go

Answer 4

sinx is bounded between -1 and 1, can u check again if its really in infinity/infinity form ?

Answer 5

yes thats true - its not in infinity/ infinity form

since the denominator is x^ (1/2) the wouldn't the limit be 0?

Answer 6

yes 0 is the answer

Answer 7

as rsadyvica said the values of sine lies between -1 and 1 and as x approaches infinity so does sqrt x

Answer 8

if you need extra help I recomend looking at the Khan academy video for this, he does a whole vid on the exact thing you need help with :)

Answer 9

ya, you guys were right--0--but I was suppose to use
squeeze theorem

Answer 10

\[\large -1\le\sin x \le 1 \implies \dfrac{-1}{1+\sqrt{x}}\le \dfrac{\sin x}{1+\sqrt{x}}\le \dfrac{1}{1+\sqrt{x}}\]

Answer 11

take the limit through out

Answer 12

show that the limit in question is bounded between 0 and 0
that proves that the limit is exactly 0