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$\begingroup$

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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    Multiply and divide by $\cos x +1$ and use $\sin^2 x+ \cos^2 x=1$.2012-09-25

4 Answers 4

1

$\lim_{x \to a} (f(x) g(x)) = \lim_{x \to a} f(x) \lim_{x \to a} g(x)$ 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.2012-09-25
5

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

4

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

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

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 $

2

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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    @NKS It is better to use `\times` instead of `*` and I have edited it in your post.2012-09-25