Limit x tends to 0 sin^2x -sin^2x/x^2-y^2|$\lim_{x\rightarrow y}\frac{\sin^2x – sin^2y}{x^2 – y^2}$| $\lim_{x\rightarrow y}\frac{\sin^2x – sin^2y}{x^2 – y^2}$-This is a very important Mathematical Problem of Limit of Class 11 of Some Important Math Books of SN Dey, RD Sharma, NCERT, WBCHSE, ISC, JEE, Etc.
Limit x tends to 0 sin^2x -sin^2x/x^2-y^2
$\lim_{x\rightarrow y}\frac{\sin^2x – sin^2y}{x^2 – y^2}$
= $\lim_{x\rightarrow y}\frac{(sin x + siny)(sin x – siny)}{(x + y)(x – y)} $
= $\lim_{x\rightarrow y}\frac{(sin x + siny)}{(x + y)}\ldotp \lim_{x\rightarrow y}\frac{(sin x – sin y)}{(x – y)}$
= $\frac{\sin y + siny}{y + y}\ldotp \lim_{x\rightarrow y}\frac{2cos\frac{x + y}{2}\ldotp sin\frac{x – y}{2}}{2.\frac{x – y}{2}} $
= $\frac{\sin y}{y}\ldotp \lim_{x\rightarrow y}cos\frac{x + y}{2}\ldotp \lim_{x\rightarrow y}\frac{sin\frac{x – y}{2}}{\frac{x – y}{2}} $
= $\frac{\sin y}{y}\ldotp cos\frac{2y}{2}\ldotp \lim_{z\rightarrow 0}\frac{\sin \frac{z}{2}}{\frac{z}{2}}$[Let,x – y = $z$ and $x\rightarrow y\Rightarrow z\rightarrow 0$]
= $\frac{\sin y}{y}\ldotp cosy\ldotp 1$
= $\frac{2sinycosy}{2y} $
= $\frac{\sin 2y}{2y}$