# Thread: Calculus I problem. Evaluating a limit.

1. ## Calculus I problem. Evaluating a limit.

$lim_{x\to0}\frac{sin\frac{1}{x}sin^2\frac{x}{2}}{x }$

Now I know I have to get x out of the denominator to get anywhere so I am not dividing by. I tried using the Quotient Law to take the limit of the numerator and denominator separately but quickly realized that I still would be dividing by zero when I go to plug zero in for x. So I guess I could use the Product Law and take two separate limits:

$lim_{x\to0}\frac{sin\frac{1}{x}}{x}lim_{x\to0}sin^ 2\frac{x}{2}$

While I CAN see that because the second part of the equation will equal zero (the sin^2(x/2) part) and thus anything multiplied by zero will equal zero I am at a loss as to further solving the left part of the equation. I once again have x in the denominator. As I mentioned before, I know that using the Quotient Law on the left part will still leave me with x in the denominator and would make it undefined, which doesn't help. I need to get x out of the denominator on the left side, in both the denominator and the (1/x) part of sin. This is where I am having trouble deciding how to go about doing that. Perhaps I shouldn't of even used the Product Law.[/quote]

2. ## Re: Calculus I problem. Evaluating a limit.

In your original limit, try replacing $x$ with the equivalent $4\cdot\frac{1}{x}\cdot\left(\frac{x}{2} \right)^2$.

3. ## Re: Calculus I problem. Evaluating a limit.

Originally Posted by jamesrb
$lim_{x\to0}\frac{sin\frac{1}{x}sin^2\frac{x}{2}}{x }$

Now I know I have to get x out of the denominator to get anywhere so I am not dividing by. I tried using the Quotient Law to take the limit of the numerator and denominator separately but quickly realized that I still would be dividing by zero when I go to plug zero in for x. So I guess I could use the Product Law and take two separate limits:

$lim_{x\to0}\frac{sin\frac{1}{x}}{x}lim_{x\to0}sin^ 2\frac{x}{2}$

While I CAN see that because the second part of the equation will equal zero (the sin^2(x/2) part) and thus anything multiplied by zero will equal zero I am at a loss as to further solving the left part of the equation. I once again have x in the denominator. As I mentioned before, I know that using the Quotient Law on the left part will still leave me with x in the denominator and would make it undefined, which doesn't help. I need to get x out of the denominator on the left side, in both the denominator and the (1/x) part of sin. This is where I am having trouble deciding how to go about doing that. Perhaps I shouldn't of even used the Product Law.
[/QUOTE]

\displaystyle \begin{align*} \lim_{x \to 0}\frac{\sin{\frac{1}{x}}\sin^2{\frac{x}{2}}}{x} &= \lim_{x \to 0}\frac{\sin{\frac{x}{2}}}{\frac{x}{2}} \cdot \lim_{x \to 0}\frac{\sin{\frac{x}{2}}}{2} \cdot \lim_{x \to 0}\sin{\frac{1}{x}} \\ &= 1 \cdot 0 \cdot \lim_{x \to 0}\sin{\frac{1}{x}} \end{align*}

Since \displaystyle \begin{align*} \sin{\frac{1}{x}} \end{align*} is a bounded function, the product of it with another function that goes to 0 will give 0.

So the limit is 0.

4. ## Re: Calculus I problem. Evaluating a limit.

$lim_{x \to 0}\frac{\sin{\frac{x}{2}}}{\frac{x}{2}} \cdot \lim_{x \to 0}\frac{\sin{\frac{x}{2}}}{2}$

Would you mind elaborating how you got to that from:
$sin^2\frac{x}{2}$

I am rusty working with trigonometric functions and identities.

5. ## Re: Calculus I problem. Evaluating a limit.

Originally Posted by jamesrb
$lim_{x \to 0}\frac{\sin{\frac{x}{2}}}{\frac{x}{2}} \cdot \lim_{x \to 0}\frac{\sin{\frac{x}{2}}}{2}$

Would you mind elaborating how you got to that from:
$sin^2\frac{x}{2}$

I am rusty working with trigonometric functions and identities.
I didn't, I got it from \displaystyle \begin{align*} \frac{ \sin^2{\frac{x}{2}} }{x} \end{align*}. Try multiplying out \displaystyle \begin{align*} \frac{\sin{\frac{x}{2}}}{\frac{x}{2}} \cdot \frac{\sin{\frac{x}{2}}}{2} \end{align*}. Do you get what I started with?

6. ## Re: Calculus I problem. Evaluating a limit.

Originally Posted by Prove It
I didn't, I got it from \displaystyle \begin{align*} \frac{ \sin^2{\frac{x}{2}} }{x} \end{align*}. Try multiplying out \displaystyle \begin{align*} \frac{\sin{\frac{x}{2}}}{\frac{x}{2}} \cdot \frac{\sin{\frac{x}{2}}}{2} \end{align*}. Do you get what I started with?
Yeah, I see what you did. Thanks.