having trouble with a Limit calculating

[akleron]

I need to calculate the limit shown in the picture.
An easy way to do so is l'hopital's rule but I cannot use it yet ( we haven't prove it in class)
There should be another way to find the Limit ( I suppose) maybe with some trigonometric identities.. but I didn't manage to figure out how.
I thought using this identity might be a good start, but didn't get much further
Help ?

 

[MarkFL]

I'd use a sum to product identity to write the limit as:

[MATH]L=-2\lim_{x\to-5}\left(\frac{\sin\left(\dfrac{x-5}{2}\right)\sin\left(\dfrac{x+5}{2}\right)}{x+5}\right)[/MATH]
And then:

[MATH]L=\lim_{x\to-5}\left(\frac{\sin\left(\dfrac{x+5}{2}\right)}{\dfrac{x+5}{2}}\right)\cdot\lim_{x\to-5}\left(\sin\left(\frac{5-x}{2}\right)\right)[/MATH]
Can you proceed?

 

[akleron]

Took my a while but I think I got it !
thanks!

 

[Dr.Peterson]

If you didn't know that identity, you might rewrite [MATH]\cos(x)[/MATH] as [MATH]\cos((x+5) - 5)[/MATH] and use the angle-difference identity. This will lead to a limit you could work out using known limits of [MATH]\frac{1-\cos(u)}{u}[/MATH] and [MATH]\frac{\sin(u)}{u}[/MATH].

 

[Steven G]

If you learned about derivatives then this is an easy problem. This limit is in the form of the alternate definition of a derivative. The limit should simply be derivative (cos(x)) at x=5 =-sin(-5)

 

[Dr.Peterson]

Good point. And another example of the importance of context to determine what is the "best" (or even expected) solution.