limit of trig ratio: lim(1 - cos5h) / (cos7h - 1) as h - 0

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lim(1-cos5h)/(cos7h-1)
h-0
find the limit
I 've tried to multiplied the top and bottom by 1-cos5h and substitute 0 in but the answer comes out 0.
The answer for this problem is -25/49[/code]
 
Are you sure the problem is correct?. Something seems wacky.
 
Re: limit of trig

Hello, Bryan!

This is a particularly messy problem . . .
\(\displaystyle \;\;\)(I assume all the variables are \(\displaystyle h\)'s.)

It uses the theorem:\(\displaystyle \L\,\lim_{\theta\to0}\frac{\sin\theta}{\theta}\:=\:1\)

Find the limit: \(\displaystyle \L\,\lim_{h\to0}\,\frac{1\,-\,\cos5h}{\cos7h\,-\,1}\)
Click to expand...

First, factor -1 from the denominator: \(\displaystyle \L\,-\frac{1\,-\,\cos5h}{1\,-\,\cos7h}\)


Multiply by \(\displaystyle \frac{1 + \cos5h}{1 + \cos5h}\cdot\frac{1+\cos7h}{1+\cos7h}\)

\(\displaystyle \L\;\;-\frac{\1\,-\,\cos5h}{1\,-\,\cos7h}\cdot\frac{1\,+\,\cos5h}{1\,+\,\cos5h}\cdot\frac{1\,+\,\cos7h}{1\,+\,\cos7h} \;= \;-\frac{(1\,-\,\cos5h)(1\,+\,\cos5h)}{(1\,-\,\cos7h)(1\,+\,\cos7h)}\cdot\frac{1\,+\,\cos7h}{1\,+\,\cos5h}\)

\(\displaystyle \L\;\;-\frac{1\,-\,\cos^25h}{1\,-\,\cos^27h}\cdot\frac{1\,+\,\cos7h}{1\,+\,\cos5h} \;=\;-\frac{\sin^25h}{\sin^27h}\cdot\frac{1\,+\,\cos7h}{1\,+\,\cos5h}\)


Multiply by \(\displaystyle \frac{25\cdot49\cdot h^2}{25\cdot49\cdot h^2}\)

\(\displaystyle \L\;\;\frac{25\cdot49\cdot h^2}{25\cdot49\cdot h^2}\cdot\frac{\sin^25h}{\sin^27h}\cdot\frac{1\,+\,\cos7h}{1\,+\,\cos5h} \;= \;-\frac{\sin^25h}{25h^2}\cdot\frac{49h^2}{\sin^27h}\cdot\frac{25}{49}\cdot\frac{1\,+\,\cos7h}{1\,+\,\cos5h}\)

\(\displaystyle \L\;\;=\;-\left(\frac{\sin5h}{5h}\right)^2\left(\frac{7h}{\sin7h}\right)^2\cdot\frac{25}{49}\cdot\frac{1\,+\,\cos7h}{1\,+\,\cos5h}\)


We note that if \(\displaystyle h\to0\), then \(\displaystyle 5h\to0\) and \(\displaystyle 7h\to0.\)

Take the limit: \(\displaystyle \L\;\lim_{h\to0}\,\left[-\left(\frac{\sin5h}{5h}\right)^2\left(\frac{7h}{\sin7h}\right)^2\cdot\frac{25}{49}\cdot\frac{1\,+\,\cos7h}{1\,+\,\cos5h}\right]\)


and we get: \(\displaystyle \L\,-(1^2)(1^2)\cdot\frac{25}{49}\cdot\frac{1 + 1}{1 + 1}\;=\;-\frac{25}{49}\;\;\) . . . ta-DAA!

 
The x in your limit threw me off. You could give ol' L'Hopital a go.

Take the derivative of numerator and denominator:

numerator...\(\displaystyle \L\\\frac{d}{dh}[1-cos(5h)]=5sin(5h)\)

denominator...\(\displaystyle \L\\\frac{d}{dh}[cos(7h)-1]=-7sin(7h)\)

You have:

\(\displaystyle \L\\\lim_{h\to\0}\frac{5sin(5h)}{-7sin(7h)}\)

Take derivative of num and den again:

numerator....\(\displaystyle \L\\\frac{d}{dh}[5sin(5h)]=25cos(5h)\)

denominator...\(\displaystyle \L\\\frac{d}{dh}[-7sin(7h)]=-49cos(7h)\)

Now, you have:

\(\displaystyle \L\\\lim_{x\to\0}\frac{25cos(5h)}{-49cos(7h)}=\frac{-25}{49}\)
 
Thanks for all the great replies
the x was a typo and i fixed it already
thanks again
 
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