# Math Help - problem with limits

1. ## problem with limits

I had some math problems that was troublesome, and I would like to know what are the steps to those problems. Here are the problems: 1)lim x->0 (tan5x/tan2x) the ans. is 5/2. 2) lim x->0 (1-cos(x))/sin(x). The ans. is 0. 3) lim x->0 (sec x-1)/(x sec(x)) the ans. is 0

2. Originally Posted by driver327
I had some math problems that was troublesome, and I would like to know what are the steps to those problems. Here are the problems: 1)lim x->0 (tan5x/tan2x) the ans. is 5/2. 2) lim x->0 (1-cos(x))/sin(x). The ans. is 0. 3) lim x->0 (sec x-1)/(x sec(x)) the ans. is 0
Problem 1]
$\frac{\tan 5x}{\tan 2x} = \frac{\sin 5x}{\sin 2x} \cdot \frac{\cos 2x}{\cos 5x}$

We can ignore the second factor because $\lim_{x\to 0}\frac{\cos 2x}{\cos 5x} = 1$ and does not change the value of the limit.

Thus, the problem reduces to find,
$\lim_{x\to 0}\frac{\sin 5x}{\sin 2x}$

Note, we can write,
$\frac{\sin 5x}{\sin 2x} = \frac{5}{2} \cdot \frac{\sin 5x}{5x} \cdot \frac{2x}{\sin 2x}$

Now, $\lim_{x\to 0}\frac{\sin 5x}{5x} = \lim_{x\to 0}\frac{\sin x}{x} =0$

Similar reasoning says that, $\lim_{x\to 0}\frac{2x}{\sin 2x} = 1$

Thus, we are left with 5/2.

3. Originally Posted by driver327
2) lim x->0 (1-cos(x))/sin(x). The ans. is 0.
Hint: Write $\frac{1-\cos x}{\sin x} = \frac{1-\cos x}{x} \cdot \frac{x}{\sin x}$

lim x->0 (sec x-1)/(x sec(x))
Hint: Multiply denominator and numerator by $\cos x$ then apply the argument above.

4. ## re: lim x->0 (sec x-1)/(x sec(x))

I did mult. the cos(x) and it got me (1-cos(x)/x) = 0. Did I do this right? One more question, is it a matter of only getting sines and cosines in the equation like this? And thanks PerfectHacker 4 helping me out.

5. Originally Posted by driver327
I did mult. the cos(x) and it got me (1-cos(x)/x) = 0. Did I do this right?
Yes
One more question, is it a matter of only getting sines and cosines in the equation like this? And thanks PerfectHacker 4 helping me out.
Usually.