Answer for the $\lim_{x\rightarrow 0} \frac{x^2}{1-\cos x}$?

Question

So I know that you can multiply by the conjugate and get the correct answer which is 2. However I wanted to know why this method gave me the wrong answer. how I tried to solve it

And also $0\times\infty$ is indeterminate. That means you need to apply a good method of finding limit. — Z Ahmed, Jun 30 '22 at 20:05
oooohhhh completely forgot that (0)^-1 is 1/0 which is undefined. thank you very much — Yinka Eromomene, Jun 30 '22 at 20:20
Maybe you could use MathJax to type your question. https://math.meta.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference — PC1, Jun 30 '22 at 21:10

score 1 · Answer 1 · edited Jun 30 '22 at 20:26

1

You could do this: \begin{align} \lim_{x\to0}\frac{x^2}{1-\cos(x)}&=\lim_{x\to0}\frac{x^2(1+\cos(x))}{1-\cos^2(x)}\\ &=\lim_{x\to0}(1+\cos(x))\frac{x^2}{\sin^2(x)}\\ &=2\lim_{x\to0}\frac{x^2}{\sin^2(x)} \end{align}

The right hand side is $(\mathrm{sinc}(x))^{-2}$, which is well known to be $1$ for $x\to0$.

edited Jun 30 '22 at 20:26

J.G.

115,835

answered Jun 30 '22 at 20:14

PC1

2,175

1

We can also write the function as $\frac{x^2}{2\sin^2(x/2)}=\frac{2}{\operatorname{sinc}^2(x/2)}$. – J.G. Jun 30 '22 at 20:35

score 0 · Answer 2 · answered Jun 30 '22 at 20:50

$0\times 0^{-1}=0 \times \frac{1}{0}$ (there's an indeterminate form)

It may solve also without L`Hopital Rule.

$\lim_{x\rightarrow 0}\frac{x^2}{1-\cos x}=\lim_{x\rightarrow 0}\frac{x^2}{1-(1-2\sin^2 \frac{x}{2})}$

$=\lim_{x\rightarrow 0}\frac{x^2}{2\sin^2 \frac{x}{2}}$

$=\lim_{x\rightarrow 0}\frac{4\times (\frac{x}{2})^2}{2\sin^2 \frac{x}{2}}$

$=2\times \lim_{x\rightarrow 0}(\frac{\frac{x}{2}}{\sin \frac{x}{2}})^2$

$=2\times 1=2$

Answer for the $\lim_{x\rightarrow 0} \frac{x^2}{1-\cos x}$?

2 Answers2