Solving Limits: $\lim_{x\rightarrowy}\frac{sin{x}-sin{y}}{x-y}

[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?

 

[HallsofIvy]

\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}}

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.

=\frac{sin{x}cos{x}-sin{y}cos{y}+sin{x}cos{y}-cos{x}sin{y}}{(x-y)(cos{x}+cos{y})}<br /> =\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})}<br /> <br /> after this i don&#039;t &#039;ve idea wht to do.Is there next idea we have to include overhere?[/QUOTE]<br /> 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:<br /> \frac{sin x cos y- cos x sin y}{x- y}= \frac{sin(x-y)}{x- y} <br /> which goes to 1 as x goes to y, and <br /> \frac{sin x cos x- sin y cos y}{x- y}<br /> which is still a problem.<br /> <br /> It would be far simpler to use L&#039;Hopital&#039;s rule. Are you allowed to do that?

 

[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?

 

[Laven]

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.

=\frac{sin{x}cos{x}-sin{y}cos{y}+sin{x}cos{y}-cos{x}sin{y}}{(x-y)(cos{x}+cos{y})}<br /> =\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})}<br /> <br /> after this i don&#039;t &#039;ve idea wht to do.Is there next idea we have to include overhere?

<br /> 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:<br /> \frac{sin x cos y- cos x sin y}{x- y}= \frac{sin(x-y)}{x- y} <br /> which goes to 1 as x goes to y, and <br /> \frac{sin x cos x- sin y cos y}{x- y}<br /> which is still a problem.<br /> <br /> It would be far simpler to use L&#039;Hopital&#039;s rule. Are you allowed to do that?[/QUOTE]<br /> <br /> Yea that&#039;s what i meant to express in latex form.Thanks for your correction.<br /> <br /> Isn&#039;t it solved by that method?[method which i did]<br /> <br /> Umm..L&#039;Hopital&#039;s rule is it the best way to solve the problem?Thanks I don&#039;t &#039;ve idea bout that but i can get it after reading this rule.If not i&#039;ll again ask you.

 

[tiny-tim]

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.

Hi Laven!

You need to learn your trigonometric identities …

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