Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

How should I attempt this question?


$$\lim_{x\to0}\frac{\cos x -1}{x^2}$$

Here are my steps. Doesn't feels correct.

$$\lim_{x\to0}\frac{\cos x -1}{x^2}=\lim_{x\to0}\frac{\cos x -1}{x}*\lim_{x\to0}\frac{1}{x}=0*\lim_{x\to0}\frac{1}{x}=0$$

share|cite|improve this question
$\lim\limits_{x\rightarrow\infty}(1/x)$ doesn't exist. You might be tempted to say it's $\pm\infty$, but $0\cdot\infty$ is an indeterminate form. Hint: you could use L'Hopital. – David Mitra Sep 25 '12 at 17:08
I can't use L'hopital. Haven't learnt that yet technically. – Yellow Skies Sep 25 '12 at 17:09
Multiply and divide by $\cos x +1$ and use $\sin^2 x+ \cos^2 x=1$. – Siminore Sep 25 '12 at 17:10
up vote 4 down vote accepted

$$\lim_{x\to0}\frac{\cos x -1}{x^2}=-\lim_{x\to0}\frac{2\sin^2\frac x 2}{x^2}=-\frac 12 \lim_{x\to0}\left(\frac{\sin\frac x 2}{\frac x 2}\right)^2$$

$$=-\frac 12 \left(\lim_{x\to0}\frac{\sin\frac x 2}{\frac x 2}\right)^2=-\frac 1 2$$

For clarity, we may put $x=2y, x\to0 \implies y=\frac x 2 \to0$

$$=-\frac 12 \left(\lim_{x\to0}\frac{\sin\frac x 2}{\frac x 2}\right)^2=-\frac 12 \left(\lim_{y\to0}\frac{\sin y}{y}\right)^2=-\frac 12 $$

share|cite|improve this answer

$$\lim_{x\to0}\frac{\cos x -1}{x^2}=\lim_{x\to0}\frac{(\cos x -1)(\cos x+1)}{x^2(\cos x+1)}$$ $$=\lim_{x\to0}\frac{\cos^2 x -1}{x^2(\cos x+1)}$$ $$=-\lim_{x\to0}\frac{\sin^2x}{x^2}\lim_{x\to0}\frac{1}{(\cos x+1)}$$ $$=-\lim_{x\to0}\left(\frac{\sin x}{x}\right)^2\lim_{x\to0}\frac{1}{(\cos x+1)}$$ $$=-\frac{1}{2}$$

share|cite|improve this answer
+1 for maximum pedagogic simplicity – Rick Decker Sep 25 '12 at 18:41

There are several mistakes in your computation, but luckily this is a very instructive counter-example! In your first step, you apply the product formula for a limit. But you can only do this when both of the factors have a limit that exists. In your case, $$ \lim_{x\to 0} \frac{1}{x} $$ does not exist (it tends to $+ \infty$), and so you can't split the limit over the product.

The second mistake you make is in trying to evaluate something of the form $0 \times \infty$. There's no way to do this: what rule do you apply - zero times anything is zero, or infinity times anything is infinity? The answer is neither, and there are examples where you have expressions of the form $0 \times \infty$ that can take any real value, or $\pm \infty$.

Other answers will give you an indication of some ways to evaluate this correctly, but given your original question, I thought I would offer some commentary on your solution.

share|cite|improve this answer
Thanks very much for your perspective! That helps. Gotta avoid those mistakes in my exams I guess. – Yellow Skies Sep 25 '12 at 17:17
@NKS It is better to use \times instead of * and I have edited it in your post. – user17762 Sep 25 '12 at 17:19

$$\lim_{x \to a} (f(x) g(x)) = \lim_{x \to a} f(x) \lim_{x \to a} g(x)$$ only when $-\infty < \lim_{x \to a} f(x), \lim_{x \to a} g(x) < \infty$.

A better way is to write the Taylor series of $\cos(x)$ as $1 - \dfrac{x^2}2 + \mathcal{O}(x^4)$. This gives us that $$\dfrac{\cos(x) - 1}{x^2} = -\dfrac12 + \mathcal{O}(x^2)$$ Hence, $$\lim_{x \to 0}\dfrac{\cos(x) - 1}{x^2} =\lim_{x \to 0}\left( -\dfrac12 + \mathcal{O}(x^2) \right) = -\dfrac12 + \lim_{x \to 0} \mathcal{O}(x^2) = -\dfrac12$$

share|cite|improve this answer
Thanks! That's pretty deep because I haven't learnt T.S yet, but I sorta get it from the first sentence. – Yellow Skies Sep 25 '12 at 17:15

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.