(1-cos 5x)/(x sin2x) limit as x tends to zero

[mathdaemon]

Hello

(1-cos 5x)/(x sin2x) limit as x tends to zero

Can you please help me solve the above limit using L'Hospital
I am stuck at (5sin 5x/(sin 2x + 2x cos2x))

Thanks

 

[HallsofIvy]

Use L'hopital's rule again.

 

[mathdaemon]

Thanks for replying

Ok so I applied L'Hospital rule again and I got
-25/4 (cos 5x/x sin 2x)
How many times do I need to apply the L'Hospital rule?
I don't think I can eliminate the x in the denominator if I keep applying in L'Hospital rule.
What do I do next?

 

[Ishuda]

Thanks for replying

Ok so I applied L'Hospital rule again and I got
-25/4 (cos 5x/x sin 2x)
How many times do I need to apply the L'Hospital rule?
I don't think I can eliminate the x in the denominator if I keep applying in L'Hospital rule.
What do I do next?

I'm not sure where you got that answer but applying L'Hospital's rule to
(5sin 5x/(sin 2x + 2x cos2x))
does not give that answer for me.

Assuming you know the answers to
limit as x -> 0 of x/sin(5x)
and
limit as x -> 0 of sin(2x)/sin(5x)
there is an easier way.

 

[mathdaemon]

Thank you for pointing it out to me. I got the answer.
I made a mistake and cancelled out two same sign terms.
Do you make silly mistakes? Can you give me some tips to avoid silly mistakes?
I know I am posting a question over question. Please close the topic if it is not allowed.

 

[Ishuda]

...Do you make silly mistakes?

Oh yeah, although if you quote me I will deny it

Can you give me some tips to avoid silly mistakes?

Sure, shoot yourself in the head at close range so you don't miss

Seriously though, just check your work, i.e. put the solution back into the original problem and see if it actually works, or see if you can solve the problem another way and get the same answer. You'll still make mistakes sometimes but hopefully it will be must less ofter.

 

[HallsofIvy]

You said "using L'Hopital's rule" so I responded to that.

Personally, I thing "L'Hopital's rule" is 'overkill'. Instead write the fraction as
\(\displaystyle \frac{1- cos(5x)}{x sin(2x)}= \frac{1- cos(5x)}{x} \frac{1}{sin(x)}\)
\(\displaystyle = 5\frac{1- cos(5x)}{5x}\frac{1}{sin(x)}\)

Now, \(\displaystyle \lim_{x\to 0}\frac{1- cos(5x)}{5x}= \lim_{y\to 0}\frac{1- cos(y)}{y}\)
where I have let y= 5x is a "standard" limit. But what about \(\displaystyle \lim_{x\to 0}\frac{1}{sin(x)}\)?

 

[Ishuda]

You said "using L'Hopital's rule" so I responded to that.

Personally, I thing "L'Hopital's rule" is 'overkill'. Instead write the fraction as
\(\displaystyle \frac{1- cos(5x)}{x sin(2x)}= \frac{1- cos(5x)}{x} \frac{1}{sin(x)}\)
\(\displaystyle = 5\frac{1- cos(5x)}{5x}\frac{1}{sin(x)}\)

Now, \(\displaystyle \lim_{x\to 0}\frac{1- cos(5x)}{5x}= \lim_{y\to 0}\frac{1- cos(y)}{y}\)
where I have let y= 5x is a "standard" limit. But what about \(\displaystyle \lim_{x\to 0}\frac{1}{sin(x)}\)?

Yes, but how does \(\displaystyle \frac{1- cos(y)}{y}\) go to zero? And we also have L'Hopital's rule is good for zero times infinity.

 

[HallsofIvy]

Yes, but how does \(\displaystyle \frac{1- cos(y)}{y}\) go to zero? And we also have L'Hopital's rule is good for zero times infinity.

Multiply both numerator and denominator by 1+ cos(y). That gives \(\displaystyle \frac{1- cos(y)}{y}= \frac{1- cos^2(y)}{y(1+ cos(y))}\)
\(\displaystyle = \frac{sin(y)}{y}\frac{sin(y)}{1+ cos(y)}\).

 

[Ishuda]

Multiply both numerator and denominator by 1+ cos(y). That gives \(\displaystyle \frac{1- cos(y)}{y}= \frac{1- cos^2(y)}{y(1+ cos(y))}\)
\(\displaystyle = \frac{sin(y)}{y}\frac{sin(y)}{1+ cos(y)}\).

The 'how does ... go to zero' was in response to your "But what about \(\displaystyle \lim_{x\to 0}\frac{1}{sin(x)}\)?"

Yes I agree you can do it a different way if you know the fundamental
for a not zero \(\displaystyle \frac{sin(ax)}{ax}\) -> 1 as x -> 0

You got down to
\(\displaystyle 5\frac{sin(y)}{y}\) \(\displaystyle \frac{sin(y)}{1+ cos(y)}\) \(\displaystyle \frac{1}{sin(2x)}\)
re-write that as
\(\displaystyle \frac{25}{2(1+ cos(y))}\)\(\displaystyle \frac{sin(5x)}{5x}\) \(\displaystyle \frac{sin(5x)}{5x}\) \(\displaystyle \frac{2x}{sin(2x)}\)
and now everything behaves nicely. I'm still not sure I wouldn't prefer L'Hospital's rule if I had to show my work.