# Math Help - How do I solve this limit?

1. ## How do I solve this limit?

$\lim_{x\to\infty}\frac{sin(x^2+5)}{x^2}$

I know that $sin(x^2+5)$ will oscillate between 1 and -1 and that $\lim_{x\to\infty}\frac{1}{x^2}$ = 0

Is this a case for using the squeeze theorem? How do I write out the answer correctly? Or is it just the limit of of the $\frac{1}{x^2}$ function = 0 and the limit of the product of two functions is the product of the limits and 0*anything = 0 ?

Thanks.

2. ## Re: How do I solve this limit?

Originally Posted by terrorsquid
$\lim_{x\to\infty}\frac{sin(x^2+5)}{x^2}$

I know that $sin(x^2+5)$ will oscillate between 1 and -1 and that $\lim_{x\to\infty}\frac{1}{x^2}$ = 0

Is this a case for using the squeeze theorem? How do I write out the answer correctly? Or is it just the limit of of the $\frac{1}{x^2}$ function = 0 and the limit of the product of two functions is the product of the limits and 0*anything = 0 ?

Thanks.
$\displaystyle \sin{(x^2 + 5)}$ is bounded between -1 and 1, and $\displaystyle x^2$ increases without bound. Therefore, whatever the value of the numerator, dividing by $\displaystyle x^2$ when $\displaystyle x$ is large makes the quotient very small.

Therefore $\displaystyle \lim_{x \to \infty}\frac{\sin{(x^2 + 5)}}{x^2} = 0$

3. ## Re: How do I solve this limit?

Maybe I over-complicated things because the question asked that I carefully justify every step of my argument. Thanks

4. ## Re: How do I solve this limit?

You could you the squeeze theorem: $-\frac{1}{x^2}\le\frac{\sin(x^2+5)}{x^2}\le\frac{1} {x^2}$. It's also easy to show this by the definition of limit since $\left|\frac{\sin(x^2+5)}{x^2}\right|\le\frac{1}{x^ 2}$ and you can find from which point $\frac{1}{x^2}<\varepsilon$.

5. ## Re: How do I solve this limit?

Originally Posted by terrorsquid
Maybe I over-complicated things because the question asked that I carefully justify every step of my argument. Thanks
Every step IS justified - it's more sophisticated to explain something in English than it is to explain something in symbols...

6. ## Re: How do I solve this limit?

Originally Posted by terrorsquid
$\lim_{x\to\infty}\frac{sin(x^2+5)}{x^2}$

I know that $sin(x^2+5)$ will oscillate between 1 and -1 and that $\lim_{x\to\infty}\frac{1}{x^2}$ = 0

Is this a case for using the squeeze theorem? How do I write out the answer correctly? Or is it just the limit of of the $\frac{1}{x^2}$ function = 0 and the limit of the product of two functions is the product of the limits and 0*anything = 0 ?

Thanks.
the trouble with the 0*anything = 0 "proof" is that the "anything" in this case, is undefined. so the "squeeze theroem" method makes more sense. or, you could appeal directly to the definition of a limit:

let $\epsilon$ > 0. choose $N > \frac{1}{\sqrt{\epsilon}}$.

then, for x > N, $\left|\frac{\sin(x^2+5)}{x^2}- 0\right| \leq \frac{1}{x^2} < \frac{1}{N^2} < \frac{1}{\frac{1}{(\sqrt{\epsilon})^2}} = \epsilon$