# limit as x goes to zero of sin(x^2)/x

• Oct 23rd 2012, 08:17 PM
kingsolomonsgrave
limit as x goes to zero of sin(x^2)/x
Attachment 25379

Does this require l'Hospitals's Rule or is there another way to do this?

l'hospital's rule would give lim x->0 cos(x^2)/1 which is = 1/1=1
• Oct 23rd 2012, 08:42 PM
MarkFL
Re: limit as x goes to zero of sin(x^2)/x
You are forgetting to apply the chain rule in the numerator...you could also multiply by x/x to get the same result of 0.
• Oct 23rd 2012, 09:10 PM
kingsolomonsgrave
Re: limit as x goes to zero of sin(x^2)/x
so for the the above it should be $\displaystyle cos(x^2)*2x/1$ which would be 1*0/1=0

multiplying by x/x we would get $\displaystyle xsin(x^2)/x^2$ which is equal is also equal to zero too but how? I feel like i'm missing some obvious step for version where we multiply by x/x

also is $\displaystyle sin(x^2)/x=sin^2x/x=(sinx)^2/x$ ? or is it just the last two that are the same?
• Oct 23rd 2012, 09:20 PM
Prove It
Re: limit as x goes to zero of sin(x^2)/x
Quote:

Originally Posted by kingsolomonsgrave
so for the the above it should be $\displaystyle cos(x^2)*2x/1$ which would be 1*0/1=0

multiplying by x/x we would get $\displaystyle xsin(x^2)/x^2$ which is equal is also equal to zero too but how? I feel like i'm missing some obvious step for version where we multiply by x/x

also is $\displaystyle sin(x^2)/x=sin^2x/x=(sinx)^2/x$ ? or is it just the last two that are the same.]?

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

Also, no, \displaystyle \displaystyle \begin{align*} \sin{\left(x^2\right)} \neq \sin^2{x} \end{align*}, rather \displaystyle \displaystyle \begin{align*} \sin^2{x} = \left(\sin{x}\right)^2 \end{align*}.
• Oct 23rd 2012, 09:20 PM
MarkFL
Re: limit as x goes to zero of sin(x^2)/x
We have:

$\displaystyle \lim_{x\to0}\frac{x\sin(x^2)}{x^2}=\lim_{x\to0}x \cdot \lim_{x\to0}\frac{\sin(x^2)}{x^2}$

For the second limit let $\displaystyle u=x^2$ and as $\displaystyle x\to0$ we have $\displaystyle u\to0$ so we may write:

$\displaystyle \lim_{x\to0}x\cdot\lim_{u\to0}\frac{\sin(u)}{u}=0 \cdot1=0$

I am assuming you have been given $\displaystyle \lim_{u\to0}\frac{\sin(u)}{u}=1$ to use for such limits.
• Oct 23rd 2012, 09:25 PM
kingsolomonsgrave
Re: limit as x goes to zero of sin(x^2)/x
Thanks, yes we were given lim as x-> 0 sin(u)/u = 1

thanks much