Limits next question

1. Jul 6, 2009

Laven

$$\lim_{x\rightarrowy}\frac{sin{x}-sin{y}}{x-y}$$

so this is the question.

I'm here solving this problem you please check where am i wrong or next idea i've to use here.
=$$\lim_{x\rightarrowy}\frac{sin{x}-sin{y}}{x-y}x\frac{cos{x}+cos{y}}{cos{x}+cos{y}}$$
=$$\frac{sin{x}cos{x}-sin{y}cos{y}+sin{x}cos{y}-cos{x}sin{y}}{{x-y}{cos{x}+cos{y}}$$
=$$\lim_{x\rightarrowy}\frac{sin{x}cos{x}-sin{y}cos{y}+sin{x-y}}{{x-y}{cos{x}+cos{y}}$$ +\fracsin{x-y}{(x-y)(cos{x}+cos{y})}[/tex]

after this i don't 've idea wht to do.Is there next idea we have to include overhere?

2. Jul 6, 2009

HallsofIvy

The "x" you had in the middle here was just "times" wasn't it? Better not to use such a symbol along with"x" as a variable.

$$=[tex]\frac{sin{x}cos{x}-sin{y}cos{y}+sin{x}cos{y}-cos{x}sin{y}}{(x-y)(cos{x}+cos{y})}$$
=$$\lim_{x\rightarrow y}\frac{sin{x}cos{x}-sin{y}cos{y}+sin{x-y}}{(x-y)(cos{x}+cos{y})} +\frac{sin(x-y)}{(x-y)(cos{x}+cos{y})}$$

after this i don't 've idea wht to do.Is there next idea we have to include overhere?[/QUOTE]
I think I have corrected your LaTex properly. sin(x- y)= sin(x)cos(y)- cos(x)sin(y) and you appear to be trying to put the numerator in that form. (It would be a good idea to explain things like that when asking about a problem.) You wind up with two parts:
$$\frac{sin x cos y- cos x sin y}{x- y}= \frac{sin(x-y)}{x- y}$$
which goes to 1 as x goes to y, and
$$\frac{sin x cos x- sin y cos y}{x- y}$$
which is still a problem.

It would be far simpler to use L'Hopital's rule. Are you allowed to do that?

3. Jul 7, 2009

Laven

$$\lim_{x\rightarrowy}\frac{sin{x}-sin{y}}{x-y}$$

so this is the question.

I'm here solving this problem you please check where am i wrong or next idea i've to use here.
=$$\lim_{x\rightarrowy}\frac{sin{x}-sin{y}}{x-y}*\frac{cos{x}+cos{y}}{cos{x}+cos{y}}$$
=$$\frac{sin{x}cos{x}-sin{y}cos{y}+sin{x}cos{y}-cos{x}sin{y}}{(x-y)(cos{x}+cos{y})}$$
=$$\lim_{x\rightarrowy}\frac{sin{x}cos{x}-sin{y}cos{y}+sin{x-y}}{(x-y)(cos{x}+cos{y})} +\fracsin{x-y}{(x-y)(cos{x}+cos{y})}$$

after this i don't 've idea wht to do.Is there next idea we have to include over
here?

4. Jul 7, 2009

Laven

I think I have corrected your LaTex properly. sin(x- y)= sin(x)cos(y)- cos(x)sin(y) and you appear to be trying to put the numerator in that form. (It would be a good idea to explain things like that when asking about a problem.) You wind up with two parts:
$$\frac{sin x cos y- cos x sin y}{x- y}= \frac{sin(x-y)}{x- y}$$
which goes to 1 as x goes to y, and
$$\frac{sin x cos x- sin y cos y}{x- y}$$
which is still a problem.

It would be far simpler to use L'Hopital's rule. Are you allowed to do that?[/QUOTE]

Yea that's what i meant to express in latex form.Thanks for your correction.

Isn't it solved by that method?[method which i did]

Umm..L'Hopital's rule is it the best way to solve the problem?Thanks I don't 've idea bout that but i can get it after reading this rule.If not i'll again ask you.

5. Jul 9, 2009

tiny-tim

Hi Laven!

You need to learn your trigonometric identities …

in this case sinx - siny = 2 sin((x - y)/2) cos((x + y)/2)