calculus limits?

Question

http://img251.imageshack.us/img251/625/segfvu0.jpg

Is there anyway I can get rid of that x beside the cos? or is the answer just 0?

Answer 1

Use the Squeeze Theorem (sometimes called Sandwich Theorem).

For any nonzero x, -1 <= cos((x + 1) / x^2) <= 1, because the cosine only outputs values between -1 and 1.

Therefore, -x <= x cos((x + 1) / x^2) <= x. (If the cosine outputs something between -1 and 1, then multiplying by x will output something between -x and x.)

Look at

-x <= x cos ((x + 1)/x^2) <= x.

The limit as x approaches 0 of -x is 0, and the limit as x approaches 0 of x is 0, so by the Squeeze Theorem, the limit as x approaches 0 of x cos ((x + 1)/x^2) is also 0.

(Intuitively: because x cos ((x + 1)/x^2) lies in between two things which are approaching 0, it must be approaching 0 as well.)

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After seeing a solution to a problem of this sort, students often ask "How would I know to use the Squeeze Theorem?" The best answer I can give is "Because it works well for this sort of problem"; if use of the Squeeze Theorem did not occur to you, here is an idea of when you should think to use it:

If you have x to some power multiplied by the sine or cosine of something, then the Squeeze Theorem will often work. For example, if you are asked to find the limit as x approaches zero of one of the following:

x sin x
x^2 cos (x)
x^3 sin (x^7 + 8000x ^2 - 4x),

you can use the squeeze theorem in each case. (What's inside the sine or cosine doesn't matter here--all that matters is we know its output is between -1 and 1.)

Although there are other limits for which the squeeze theorem works, limits of this type-- x to the something times sine or cosine of something--are what you encounter most commonly in a Calculus course.

Answer 2

This limit is actually 0. We have lim (x --> 0) x cos(x/(1 + x^2)). Cos is a bounded function, its values are in [-1, 1]. So, for every x, |x cos(x/(1 + x^2))| = |x| |cos(x/(1 + x^2))| <= |x| 1 = |x|. So, when x-->0, cos(x/(1 + x^2)) --> 0.