Evaluating $\displaystyle\lim_{x\rightarrow y}\frac{\sin^2x-\sin^2y}{x^2-y^2}$

Question

I have to evaluate $\displaystyle\lim_{x\rightarrow y}\frac{\sin^2x-\sin^2y}{x^2-y^2}$

I tried several ways like replacing $x^2-y^2$ by $z$ and then solving and breaking the denominator into $(x+y)(x-y)$ and then replacing $(x-y)$ by $z$ and then solving but always I end up with complicated things which are nowhere near the answer.

How do I solve?

Thanks for any help!!

Hint: $\lim_{x \to y} (\sin^2 x - \sin^2 y)/(x-y)$ is the definition of the derivative of $,\cdots,$ at $,\cdots,$ — dxiv, Dec 14 '16 at 18:56
@MatthewLeingang You may use...but a more elementary solution is expected.... — Soham, Dec 14 '16 at 18:58

score 11 · Accepted Answer · edited Dec 14 '16 at 19:10

11

$$\begin{align*}\lim_{x\to y} \frac{\sin^2(x)-\sin^2(y)}{x^2-y^2}&=\lim_{x\to y}\frac{\sin x-\sin y}{x-y}\frac{\sin x+\sin y}{x+y}\\&=\frac{2\sin y}{2y}\lim_{x\to y}\frac{\sin x-\sin y}{x-y}\\&=\frac{\sin y}{y}(\sin y)'\\&=\frac{\sin y \cos y}{y}\end{align*}$$

$$=\frac{\sin2y}{2y}$$

edited Dec 14 '16 at 19:10

Soham

9,990

answered Dec 14 '16 at 18:59

Have you used L'Hopital?[Actually I am not too familiar with it...so can you be a little more explanatory?] – Soham Dec 14 '16 at 19:02
@tatan I use the definition of the derivative – Dec 14 '16 at 19:03
Thanks!!!!This was helpful!!(+1)... – Soham Dec 14 '16 at 19:08
1

$=\sin2y/2y$. I think it looks better that way. – Simply Beautiful Art Dec 14 '16 at 19:09

haqnatural · Answer 2 · 2016-12-14T19:04:13.217

$$\lim _{ x\rightarrow y } \frac { \sin ^{ 2 } x-\sin ^{ 2 } y }{ x^{ 2 }-y^{ 2 } } =\lim _{ x\rightarrow y } \frac { \left( \sin { x-\sin { y } } \right) \left( \sin { x+\sin { y } } \right) }{ \left( x-y \right) \left( x+y \right) } =\\ =\lim _{ x\rightarrow y } \frac { \left( 2\sin { \frac { x-y }{ 2 } } \cos { \frac { x+y }{ 2 } } \right) \left( \sin { x+\sin { y } } \right) }{ \left( x-y \right) \left( x+y \right) } =\\ =\lim _{ x\rightarrow y } \frac { \sin { \frac { x-y }{ 2 } } }{ \frac { \left( x-y \right) }{ 2 } } \frac { \cos { \frac { x+y }{ 2 } } \left( \sin { x+\sin { y } } \right) }{ \left( x+y \right) } =\\ =\cos { y } \frac { 2\sin { y } }{ 2y } =\frac { \sin { y\cos { y } } }{ y } $$

score 2 · Answer 3 · answered Dec 14 '16 at 18:58

2

Take $x=y+h$

$$\lim_{h\to0}{\sin^2(y+h)-\sin^2y\over (y+h)^2-y^2}=\lim_{h\to0}{\sin(y+h)-\sin(y)\over h}\cdot{\sin(y+h)+\sin(y)\over 2y+h}\\=\cos(y)\cdot{2\sin(y)\over 2y}=\cos(y)\cdot{\sin(y)\over y}$$

answered Dec 14 '16 at 18:58

Qwerty

6,165

score 2 · Answer 4 · answered Dec 14 '16 at 18:59

2

$\lim_\limits{x\to y} \frac{\sin^2 x - \sin^2 y}{x^2-y^2} =\lim_\limits{x\to y} \left(\frac{\sin x - \sin y}{x-y}\right)\left(\frac{\sin x + \sin y}{x+y}\right)\\ (\frac {d}{dy} \sin y)(\frac {\sin y}{y})$

answered Dec 14 '16 at 18:59

Doug M

57,877

score 1 · Answer 5 · answered Dec 14 '16 at 19:07

Use the following formulas

$$\begin{align}\sin{p}+\sin{q}=&2\sin{p+q\over 2}\cos{p-q\over 2}\\ \sin{p}-\sin{q}=&2\sin{p-q\over 2}\cos{p+q\over 2} \end{align}$$

to rewrite

$${\sin^2{x}-\sin^2{y}\over x^2-y^2}=2{\sin{x-y\over 2}\over{x-y\over 2}}{\sin{x+y\over 2}\over x+y}\cos{x+y\over 2}\cos{x-y\over 2}$$

And so the limit is

$${\sin{y}\cos{y}\over y}$$

Because ${\sin{x-y\over 2}\over {x-y\over 2}}\to 1$ (the rest is straightforward)

score 1 · Answer 6 · answered Dec 14 '16 at 19:13

$\lim\limits_{x\rightarrow y}\frac{\sin^2x-\sin^2y}{x^2-y^2} = \lim\limits_{x\rightarrow y}\frac{\sin^2x-\sin^2y}{x-y}\frac{1}{x+y} = \left(\frac{d}{d y}\sin^2y\right)\frac{1}{y+y} = \frac{2\sin y\cos y}{2y} = \frac{\sin y\cos y}{y}$

It can also be written a few other ways, as

$\frac{\sin y \cos y}{y} = \frac{\sin2y}{2y} = \text{sinc} \ {2y}$

Evaluating $\displaystyle\lim_{x\rightarrow y}\frac{\sin^2x-\sin^2y}{x^2-y^2}$

6 Answers6