# Math Help - Limit problem

1. ## Limit problem

I expanded everything out and got [3x^2 + x^2*sin(x)] / [x^2 + 2sin(x) + sin^2(x)] and I tried splitting the problem but it didn't work because the denominator is always 0 when I try plugging in '0' for 'x.' I can't find any way to cancel out the denominator

2. ## Re: Limit problem

Originally Posted by PhizKid

I expanded everything out and got [3x^2 + x^2*sin(x)] / [x^2 + 2sin(x) + sin^2(x)] and I tried splitting the problem but it didn't work because the denominator is always 0 when I try plugging in '0' for 'x.' I can't find any way to cancel out the denominator
When you expand the denominator you get

$x^2+2x\sin(x)+\sin^2(x)=x^2\left[1+2\frac{\sin(x)}{x}+\left( \frac{\sin(x)}{x}\right)^2\right]$

Now reduce.

P.S. you should know the value of the limit

$\lim_{x \to 0}\frac{\sin(x)}{x}$

Can you finish from here?

3. ## Re: Limit problem

With this kind of limit, I like to first note which factors are relevant. That (3+sin(x)) part isn't - it's going to 3, so isn't going to be the cause of the limit converging or not. The other two factors are what count - the x-squared in the numerator, and the (x+sin(x)) squared in the demoniator, are what is making this be of the form 0/0; that's where the action is; that's where you need to pull some trick or simplification to find the limit.

As a hint, when it's "like 0/0", and has x's and sin(x)'s and is a limiti as x goes to 0, almost surely you'll end up using sin(x)/x goes to 1 as x goes to 0. You could even do the following:
"Replace" sin(x) by (sin(x)/x)x, and then keep that (sin(x)/x) together in your algebra, as it'll be going to 1. Now you don't have anymore sin(x)'s running around - only x's and that (sin(x)/x) that's going to 1.

For this problem, a big hint would be to observe that:

$\frac{x^2}{(x+\sin(x))^2} = \left(\frac{x}{x+\sin(x)}\right)^2 = \left(\frac{1}{1+\frac{\sin(x)}{x}}\right)^2$

4. ## Re: Limit problem

Originally Posted by johnsomeone
With this kind of limit, I like to first note which factors are relevant. That (3+sin(x)) part isn't - it's going to 3, so isn't going to be the cause of the limit converging or not. The other two factors are what count. As a hint, when it's "like 0/0", and has x's and sin(x)'s and is a limiti as x goes to 0, almost surely you'll end up using sin(x)/x goes to 1 as x goes to 0.

As a hint, note that
$\frac{x^2}{(x+\sin(x))^2} = \left(\frac{x}{x+\sin(x)}\right)^2 = \left(\frac{1}{1+\frac{\sin(x)}{x}}\right)^2$
I don't understand how you went from $\left(\frac{x}{x+\sin(x)}\right)^2 = \left(\frac{1}{1+\frac{\sin(x)}{x}}\right)^2$
This was the hint provided in the textbook, but I failed see how the numbers have suddenly all changed during this step.

5. ## Re: Limit problem

$\frac{x}{x+\sin(x)}$

$= \left(\frac{x}{x+\sin(x)}\right) \ \frac{(1/x)}{(1/x)}$

$= \frac{(x)(1/x)}{(x+\sin(x))(1/x)}$

$= \frac{1}{(x)(1/x) +(\sin(x))(1/x)}$

$= \frac{1}{1+\frac{\sin(x)}{x}}$

6. ## Re: Limit problem

Originally Posted by PhizKid
I don't understand how you went from $\left(\frac{x}{x+\sin(x)}\right)^2 = \left(\frac{1}{1+\frac{\sin(x)}{x}}\right)^2$
This was the hint provided in the textbook, but I failed see how the numbers have suddenly all changed during this step.
Are you saying that you don't understand how we divide numerator and denominator by $x~?$

7. ## Re: Limit problem

Oh, I see that now. But how did you know to divide both the numerator and denominator by x?

8. ## Re: Limit problem

Originally Posted by PhizKid
Oh, I see that now. But how did you know to divide both the numerator and denominator by x?
Before I wrote "As a hint, when it's "like 0/0", and has x's and sin(x)'s and is a limiti as x goes to 0, almost surely you'll end up using sin(x)/x goes to 1 as x goes to 0.".
When I see something like this problem, after deciding that the sin(x) going to 0 is an essential reason that the limit's value is unclear, I think to myself: "I need to get an x under that sin(x)." In this case, the way to do that was to divide both numerator and demoninator by x.