# How to solve $\lim_{x\to\infty}\frac{cos x}{x-1}$?

How to solve the following question?

\begin{eqnarray} \\\lim_{x\to\infty}f(x)&=&\lim_{x\to\infty}\frac{cos x}{x-1}\\ \\&=&\lim_{x\to\infty} \frac{\sqrt{1-sin^2x}}{x-1}\\ \\&=&\lim_{x\to\infty} \frac{ \frac{\sqrt{1-sin^2x}}{x}}{\frac{x-1}{x}}\\ \\&=&\lim_{x\to\infty} \frac{\sqrt{\frac{1}{x^2}-1}}{1-\frac{1}{x}}\\ \\&=&\frac {\sqrt{-1}}{1}\\ \\&=&Error Math\\ \end{eqnarray}

The ans is 0.

• Sorry, my mistake. It should be $\lim_{x\to\infty}$ instead of $\lim_{x\to0}$. – Kin Oct 18 '13 at 14:15
• $\cos x$ stays between $-1$ and $1$, and $x-1$ gets big. – André Nicolas Oct 18 '13 at 14:16
• @CasperLI : I think your false is that you replace $\frac{sin^2(x)}{x^2}$ with 1 , it applies when $x \to 0$ – chaviaras michalis Aug 29 '16 at 17:03

As $x\rightarrow \infty$ the fraction $\frac{1}{x-1}\rightarrow 0$

Since $|cos(x)|\leq 1$ the limit $\frac{cos(x)}{x-1}\rightarrow 0$ as $x\rightarrow \infty$.

• Since cos(x) is in [-1,1] and the denominator tend to infinity, the result should be 0? – Kin Oct 18 '13 at 14:34
• @Casper Li Exactly! – epsilon Oct 18 '13 at 20:47

This can be solved by using the Sandwich Theorem.

You know that:

$$-1 ≤ \cos x ≤ 1$$

Therefore the function $\frac{\cos x}{x - 1}$ will have the range of:

$$[\frac{-1}{x - 1}, \frac{1}{x-1}]$$

Apply the limit to this range, what do you get?

This is a result of the fact that cos(x) is a bounded function.

Well, if you want the first place where your solution goes wrong, it is in this equality that you use: $\cos x = \sqrt{1 - \sin^2 x}$. This is simply not true.

On an unrelated note, you don't really need to make any smart substitutions for $\cos x$. To find the limit it is enough to know that $\cos x$ is a bounded function.

• This is indeed the first error, but not the most important one (the extra modulus doesn't change much). The next transition goes really wrong, though. – Jakub Konieczny Oct 18 '13 at 14:19
• $cosx = \sqrt{cos^2x} = \sqrt{1-sin^2x}$ I cannot do that? – Kin Oct 18 '13 at 14:27
• Yes, you cannot. You use the incorrect identity $x = \sqrt{x^2}$. In reality, $|x| = \sqrt{x^2}$. – Dan Shved Oct 18 '13 at 14:28
• No, because $\cos x$ has negative values too, which cannot be reached by a square-root. :) – mikhailcazi Oct 18 '13 at 14:28
• This doesn't mean that the original identity from which you've derived this is incorrect too. $\cos^2x + \sin^2x = 1$ is perfectly applicable. – mikhailcazi Oct 18 '13 at 14:30

I don't see how you pass from $\sqrt{1-\sin ^2 x}$ to $\sqrt{1/x^2-1}$ (the latter is obviously ill defined for $x \to \infty$. If you are trying to use $\sin x \sim x$, this only holds for small $x$.

The limit should be $0$ because $1/(x-1) \to 0$ and $\cos$ is bounded.

You should not substitute $x$ instead of $\sin x$.