# Limit of and Sin(9x)/x and 1/Cos(9x)

1. Jan 25, 2016

### Staff: Mentor

In a homework problem I had to find the limit as x goes to 0 of the function: sin(7x)/[x+tan(9x)]

Substituting sin(9x)/cos(9x) in for tan(9x) then dividing the top and bottom by x and finding the limit supposedly yields 7/1+(9)(1), giving an answer of 7/10.

What I don't get is why the limit as x goes to 0 of sin(9x)/x is 9, but the limit of 1/cos(9x) is 1 and not 1/9. Would it make a difference if it was x/cos(9x) instead?

2. Jan 25, 2016

### axmls

Because $\cos(9x)$ is not $0$ at $x = 0$, you can simply plug in the value of $0$ to see that $\lim_{x \to 0} 1/\cos(9x) = 1/1 = 1$. We can't do this for the $\sin(9x)/x$ because it is not defined for $x = 0$.

Likewise, for $\lim_{x \to 0} x/\cos(9x)$, we have $\lim_{x \to 0} x/\cos(9x) = 0/1 = 0$

If you want an intuitive reason, it's because for $x$ close to $0$, you may have learned that $\sin(x) \approx x$. Then it makes sense that $\lim_{x \to 0} \sin(9x)/x = 9$.

If you know L'Hôpital's rule, then this can also be seen, and there is also a geometric proof using the squeeze theorem to show the correct limit for your sine example, but honestly, the best thing to do is look at some graphs of the two.

Another intuitive approach: the cosine function is approaching $1$ as $x \to 0$, so the ratio of $\frac{k}{\cos(x)}$ as $x \to 0$ will just be $k$ (there are no problems with dividing by zero.
Things become more tricky with the sine function, because it is approaching 0. If it were some constant k divided by the sine function, you can see that the limit would not exist, but because both the numerator and denominator are approaching 0, it's a bit more subtle than that, because it's possible for both of them to be approaching in such a way that they have a constant ratio for $x$ near $0$.

If you have not covered L'Hôpital's rule yet, it will make more sense when you do.

Last edited: Jan 25, 2016
3. Jan 25, 2016

### Staff: Mentor

Okay, that makes much more sense.

Ah, I see it now. Thanks!

4. Jan 28, 2016

### axmls

I was working a similar problem with someone today and I realized I forgot to point this out, if you haven't already done it this way.

Consider the following limit: $$\lim_{x \to 0} \frac{\sin(k x)}{x}$$
Where $k$ is a nonzero constant. Then we can take advantage of the fact that we know $$\lim_{x \to 0} \frac{\sin(x)}{x} = 1$$
by making the substitution $u = kx$, so that we get
$$\lim_{u \to 0}\frac{\sin(u)}{\frac{u}{k}}=\lim_{u \to 0} k \frac{\sin(u)}{u} = k \lim_{u \to 0} \frac{\sin(u)}{u} = k \cdot 1 = k$$
If you haven't seen it done that way, I feel like it makes it a bit more intuitive.