Re: limits help please!!!

Quote:

Originally Posted by

**ubhutto** i don't know how to prove the following limits

limit of ( 1 - cos x) / x^2 when x is +infinity

limit of (tan4x) / 6 when x is 0

limit of (sin x^2) / x

in the first equation i got to the point of multiplying by the reciprocal but i don't know what to do next

in the second equation i converted tan to sin/cos but i dont know where to go from there

In the first, $\displaystyle \displaystyle \begin{align*} 1 - \cos{x} \end{align*}$ is bounded between 0 and 2, and the denominator goes to $\displaystyle \displaystyle \begin{align*} \infty \end{align*}$, which makes the fraction go to 0.

For the second, surely you can just substitute in x = 0.

For the third, what is the value of x going to?

Re: limits help please!!!

Re: limits help please!!!

Well multiply top and bottom by x to give

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

Can you go from there?

Re: limits help please!!!

i dont understand what to do with the sinx^2 / x^2 , how do i simplify it, if you can help me with that part ill be able to do the rest of the problems i have

Re: limits help please!!!

Quote:

Originally Posted by

**ubhutto** i dont understand what to do with the sinx^2 / x^2 , how do i simplify it, if you can help me with that part ill be able to do the rest of the problems i have

You should know that $\displaystyle \displaystyle \begin{align*} \lim_{X \to 0}\frac{\sin{X}}{X} = 1 \end{align*}$. Here your $\displaystyle \displaystyle \begin{align*} X \end{align*}$ just happens to be $\displaystyle \displaystyle \begin{align*} x^2 \end{align*}$.