# Limit problem solution

1. Mar 24, 2009

### icosane

1. The problem statement, all variables and given/known data

lim as x goes to 0 of

[x*csc(2x)] / cos(5x)

3. The attempt at a solution

The book solution says the answer is 1/2. I keep getting zero as an answer because of the x in the numerator and am unsure of how else to go about the problem. I'm pretty sure I'm supposed to use the limit as x goes to 0 of sinx / x, but even when I get to that point the limit of x as x goes to 0 always gives me an answer of zero. What am I doing wrong? Thanks!

2. Mar 24, 2009

### Staff: Mentor

The part to be concerned with is x*csc(2x), since the limit of cos(5x) is 1, as x goes to 0. If you can establish a limit for x*csc(2x), then lim x*csc(2x)/cos(5x) will be = lim x*csc(2x).

x*csc(2x) = x/sin(2x), so all you need now is 2x in the numerator instead of the x that is there. Can you turn x into 2x by multiplying by 1 in some form?

3. Mar 24, 2009

### lanedance

so you have

$$\frac{x.csc(2x)}{cos(5x)} = \frac{x.}{sin(2x).cos(5x)}$$

so the denmoinator goes to zero as well, so you can't say it is zero...

4. Mar 24, 2009

### icosane

So the lim as x goes to 0 of x/sin(x) is also equal to 1? I assume that is the case because multiplying by 2/2 gives the solution of 1/2. Thanks a lot, I really appreciate it.

Just out of curiosity now, is the lim as x goes to 0 of x/(1-cos(x)) = infinity?

5. Mar 24, 2009

### lanedance

lim as x goes to 0 sin(x)/x is one, as is sin(x)/x

$$\stackrel{lim}{x \rightarrow 0} f(x) = c$$

then what is
$$\stackrel{lim}{x \rightarrow 0} \frac{1}{f(x)} = ?$$

the second limit is true
$$\stackrel{lim}{x \rightarrow 0} \frac{x}{(1-cos(x))} = \infty$$
can you show why?

(l'hopitals rule is good for all of these if you know it...)

6. Mar 25, 2009

### Staff: Mentor

No, the limit doesn't exist.
$$\lim_{x \rightarrow 0^+} x/(1 -cos(x)) = \infty$$
while

$$\lim_{x \rightarrow 0^-} x/(1 -cos(x)) = -\infty$$

7. Mar 25, 2009

### lanedance

good pickup - my mistake

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