Limit Problem Solution: Solving [x*csc(2x)] / cos(5x) as x approaches 0

[icosane]

Homework Statement

lim as x goes to 0 of

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

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!

 

[Mark44]

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?

 

[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...

 

[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?

 

[lanedance]

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

think about this, if
\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...)

 

[Mark44]

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?

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

 

[lanedance]

good pickup - my mistake