• Oct 1st 2012, 04:58 PM
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
• Oct 1st 2012, 05:23 PM
Prove It
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 \begin{align*} 1 - \cos{x} \end{align*} is bounded between 0 and 2, and the denominator goes to \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?
• Oct 1st 2012, 05:29 PM
ubhutto
it is going to 0
• Oct 1st 2012, 05:33 PM
Prove It
Well multiply top and bottom by x to give

\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?
• Oct 1st 2012, 05:39 PM
ubhutto
You should know that \displaystyle \begin{align*} \lim_{X \to 0}\frac{\sin{X}}{X} = 1 \end{align*}. Here your \displaystyle \begin{align*} X \end{align*} just happens to be \displaystyle \begin{align*} x^2 \end{align*}.