Question

How do I compute this limit using the Squeeze Theorem?: http://www.wolframalpha.com/input/?i=lim+x-%3E0+(8x%5E4+*+sin(8%2Fx)+%2B+6cos(x))%2F(8+%2B+7+*+sqrt(1%2Bx%5E4))

I can actually do this in my head lol but I can't figure out how to do use the Squeeze Theorem to do it.

For typical problems, I know to compute the limits of the leftmost and rightmost parts of the following inequality, that is, f(x) and g(x).

f(x) <= h(x) <= g(x)

with

lim x->x_0 f(x) = L = lim x->x_0 g(x)

such that

lim x->x_0 h(x) = L

but I am confused as to how I should go about this problem.

Answer 1

i think that this problem needs to be broken into parts and that you have to use the squeeze theorem for the sin(x) and cos function since the limit of a sum is the sum of the limits. So you can start the sin function by -1<sin(8/x)<1. then you multiply all the terms by 8x^4 and you will see that the limit on either side goes to 0 so the limit of the middle term also goes to zero. try again for the cos function the same tactic.

Answer 2

I don't think I need to do the same for cos(x) but thanks to you, I feel I get how to do it (unless I misunderstood you and this is what you meant all along).

If you could confirm what I did below is correct, that would be great :) .:

-1 <= sin(8/x) <= 1

-8x^4 <= 8x^4 * sin(8/x) <= 8x^4

-8x^4 + 6cos(x) = 8x^4 * sin(8/x) + 6cos(x) <= 8x^4 + 6cos(x)

(-8x^4 + 6cos(x))/(8+7sqrt(1+x^4)) <= (8x^4 * sin(8/x) + 6cos(x))/(8+7sqrt(1+x^4)) <= (8x^4 + 6cos(x))/(8+7sqrt(1+x^4))

Here is the last step (the inequality above) in Wolfram Alpha since it looks better: http://www.wolframalpha.com/input/?i=%28-8x%5E4+%2B+6cos%28x%29%29%2F%288%2B7sqrt%281%2Bx%5E4%29%29+%3C%3D+%288x%5E4+*+sin%288%2Fx%29+%2B+6cos%28x%29%29%2F%288%2B7sqrt%281%2Bx%5E4%29%29+%3C%3D+%288x%5E4+%2B+6cos%28x%29%29%2F%288%2B7sqrt%281%2Bx%5E4%29%29

Answer 3

yes that it is what i would do.

Answer 4

Thanks for the confirmation! :)