# What mistake have I committed in doing these steps to find the limit?

How should I attempt this question?

Find

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

Here are my steps. Doesn't feels correct.

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

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$\lim\limits_{x\rightarrow\infty}(1/x)$ 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

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

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

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

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

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$$\lim_{x \to a} (f(x) g(x)) = \lim_{x \to a} f(x) \lim_{x \to a} g(x)$$ 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, $$\lim_{x \to 0}\dfrac{\cos(x) - 1}{x^2} =\lim_{x \to 0}\left( -\dfrac12 + \mathcal{O}(x^2) \right) = -\dfrac12 + \lim_{x \to 0} \mathcal{O}(x^2) = -\dfrac12$$

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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

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.

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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

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

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+1 for maximum pedagogic simplicity – Rick Decker Sep 25 '12 at 18:41