Question

how about the lim x-> infinity of (sin^2)(x)/(x^2+1) do i also multiply by 1/x^2?

Answer 1

does this has to do with finding horizontal aysmptoe?

Answer 2

yea

Answer 3

the answer is zero

Answer 4

is it because both top and bottom go to zero when you evaluate?

Answer 5

Its because the power of the denominator is greater than the power of the numerator. x^2 grows faster than x so the function keeps becoming smaller and smaller as x ---> infinity.

Answer 6

is the numerator supposed to be \(\sin^2(x)\)?

Answer 7

yea!

Answer 8

For rational functions, if the power of the numerator is greater, the limit goes to +/- infinity depending on lead coefficient's sign. If the power of the demoninator is greater, the limit goes to 0. If the power of the numerator = power of denominator, the limit as x ---> infinity is the quotients of the lead coefficients. So for example:\[\lim_{x \rightarrow \infty} \frac {x^5 - 4x^2 +1}{x^3 - 1} = \infty\]\[\lim_{x \rightarrow \infty} \frac {x^4 - \sin x + 2}{x^9 - 7x + 1} = 0\]\[\lim_{x \rightarrow \infty} \frac {-7x^6 + 4x^2 + 4x - 1}{9x^6 - \sin^2 2x + 1} = \frac {-7}{9}\]

Answer 9

then \[0\le\frac{\sin^2(x)}{x^2+1}\le\frac{1}{x^2+1}\to 0\text{ as }x\to\infty\]