Question

limit as x goes to infinity (x^2+sinx)/(x^2+1)

Answer 1

It's 1 I think. Break the limit as: \[\lim_{x \rightarrow \infty} \frac{x^2}{x^2 + 1} + \lim_{x \rightarrow \infty} \frac{sinx}{x^2 +1} \]

Answer 2

Then for the first one, apply L'Hopital's Rule. For the second one, I think you will have to use the sandwich theorem.

Answer 3

But it will be zero.

Answer 4

it is 1 but how is sinx 0? when it approaches inf

Answer 5

I mean, the second one will be 0.

Answer 6

\[\lim_{x \rightarrow \infty} \frac{x^2 + sinx}{x^2 + 1}\] Since the highest power in the numerator is equal to the highest power in the denominator, you just write the leading coeffecients of the highest powers as a ratio and thats the limit . So you get 1/1 so the limit is 1

Answer 7

No, sinx is limited [-1,1], right? But (x^2 + 1) in the denominator is unbounded. So, it will go to infinity, and the limit will go to zero.

Answer 8

lol didnt even realize that.... lol you can l'hop it because its not 0/0 or inf/inf lol

Answer 9

ahhh you could also do it that way I had totally forgot thanks for the reminder tyler, mathematically you have to divide by x^2 on top and bottom

Answer 10

If you need this with more rigor, you will have to apply the sanwich theorem for the sinx. But most teachers accept what I wrote above, for the unboundedness of the denominator.

Answer 11

yeah what you wrote would suffice for my class thank you bmp

Answer 12

No problem :-)