Question

$lim \: x \: to \: y \: \frac{ {sin}^{2} x - {sin}^{2} y}{ {x}^{2} - {y}^{2} }$
Evaluate the above please​

Answer 1

$\large\underline{\sf{Solution-}}$

Given expression is

$\rm :\longmapsto\:\displaystyle\lim_{x \to y} \frac{ {sin}^{2}x - {sin}^{2}y}{ {x}^{2} - {y}^{2} }$

If we substitute directly x = y, we get

$\rm \:  =  \: \dfrac{ {sin}^{2}x - {sin}^{2} x}{ {x}^{2} - {x}^{2} }$

$\rm \:  =  \: \dfrac{0}{0}$

which is indeterminant form.

So, to solve it further,

$\rm :\longmapsto\:\displaystyle\lim_{x \to y} \frac{ {sin}^{2}x - {sin}^{2}y}{ {x}^{2} - {y}^{2} }$

$\rm \:  =  \: \displaystyle\lim_{x \to y} \dfrac{sin(x + y)sin(x - y)}{(x + y)(x - y)}$

can be rewritten as

$\rm \:  =  \: \displaystyle\lim_{x \to y} \frac{sin(x + y)}{x + y} \times \displaystyle\lim_{x \to y} \frac{sin(x - y)}{x - y}$

$\rm \:  =  \: \dfrac{sin(y+ y)}{y + y} \times \displaystyle\lim_{x - y \to 0} \frac{sin(x - y)}{x - y}$

We know,

$\boxed{ \tt{ \: \displaystyle\lim_{x \to 0} \frac{sinx}{x} = 1 \: }}$

So, using this, we get

$\rm \:  =  \: \dfrac{sin2y}{2y} \times 1$

$\rm \:  =  \: \dfrac{sin2y}{2y}$

Hence,

$\rm \implies\:\boxed{ \tt{ \: \displaystyle\lim_{x \to y} \frac{ {sin}^{2}x - {sin}^{2}y}{ {x}^{2} - {y}^{2} } = \frac{sin2y}{2y} \: }}$

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More to Know :-

$\boxed{ \tt{ \: \displaystyle\lim_{x \to 0} \frac{sinx}{x} = 1 \: }}$

$\boxed{ \tt{ \: \displaystyle\lim_{x \to 0} \frac{tanx}{x} = 1 \: }}$

$\boxed{ \tt{ \: \displaystyle\lim_{x \to 0} \frac{log(1 + x)}{x} = 1 \: }}$

$\boxed{ \tt{ \: \displaystyle\lim_{x \to 0} \frac{ {e}^{x} - 1}{x} = 1 \: }}$

$\boxed{ \tt{ \: \displaystyle\lim_{x \to 0} \frac{ {a}^{x} - 1}{x} = loga \: }}$

Answer 2

Answer:

$\huge \pink{ \frac{ \sin(y) \cos(y) }{y} }$

Step-by-step explanation:

Note that in given expression y is not variable . Think of y as some constant .

As we can see it's 0/0 form . To avoid confusion let's change it into some familiar form

Now let

$x - y = z \\ \implies \: x = y + z$

Now our limit changes into

$\displaystyle{ \lim_{z \to0} \frac{ \sin {}^{2} (y + z) - \sin {}^{2} (y) }{(y + z) {}^{2} - {y}^{2} } } \\ \\ \implies \\ \displaystyle{ \lim_{z \to0} \frac{ \sin (y + z) - \sin (y) }{z} } .\frac{ \sin (y + z) + \sin (y) }{2y + z}$

Now there are two terms in multiply first term is 0/0 form so differentiate it wrt z ( remember y is just a constant )

$\displaystyle{ \lim_{z \to0} \frac{ \cos(y + z) - 0 }{1} } .\frac{ \sin (y + z) + \sin (y) }{2y + z}$

Now put limit z to 0

$\frac{ \cos(y + 0). \{\sin(y + 0) + \sin(y) \} }{(2y + 0)} \\ \\ = \frac{ \cos(y). 2\sin(y) }{2y} \\ \\ = \frac{ \sin(y) \cos(y) }{y}$

Note that it's same answers as mathdude