Limit (sinx)^2/x^2 answer check please

[synergix]

Homework Statement

Limit sin²x/x²
x->infinity

The Attempt at a Solution

0 because sin will never go above 1 or below -1 and x² will approach infinity

am I right??

 

[Dick]

Right. Formally, you conclude the limit is 0 from the squeeze theorem.

 

[synergix]

Thanks. But I never understood the squeeze theorem it was never explained to me very well.

 

[Dick]

You are saying 0<=sin(x)^2/x^2<=1/x^2 (sin(x) can be between -1 and 1 but 0<=sin(x)^2<=1 since it's squared). Since the two outside limits are zero, lim sin(x)^2/x^2 must also be zero. It's 'squeezed' between the two outside limits.

 

[NastyAccident]

Homework Statement

Limit sin²x/x²
x->infinity

The Attempt at a Solution

0 because sin will never go above 1 or below -1 and x² will approach infinity

am I right??

Personally, I start by using the comparison theorem for these types of limits/integrals. This helps me figure out whether it will converge or diverge and also gives me the limit.

sin(x) is always going to be less than or equal to 1 and greater than or equal to -1. Same goes for cos(x).

sec(x) or csc(x) will always be greater than or equal to one.

So, consider:

lim sin^2 (x)/x^2 < lim 1/x^2
x->infinity x->infinity

limit 1/x^2
x->infinity

limit 0
x->infinity

So, sin^2(x)/x^2 converges and its limit is 0.

*note* I'm a Calc II Student. *note*

 

[Landau]

Or you refer to the standard limit $$\lim_{x\to\infty}\frac{\sin x}{x}=0$$ and use the product rule. Of course this 'standard limit' can be proved with the squeezing theorem, so if you're not allowed to use it (as 'standard') you might as well follow Dick.