Question

sinx-siny divided by x-y in limits​

Answer 1

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Answer 2

The value of the limit is equal to cos(y).

Given:

$\lim_{x\to y} \frac{sinx-siny}{x-y}$

To Find:

The value of the limit:   $\lim_{x\to y} \frac{sinx-siny}{x-y}$

Solution:

→ We would evaluate the limit by simplifying it step-by-step:

• Using the formula:  $sinA-sinB=2cos(\frac{A+B}{2}) sin(\frac{A-B}{2} )$

→ We get:

                       $Limit=\lim_{x\to y} \frac{1}{x-y}[ 2cos(\frac{x+y}{2} )sin(\frac{x-y}{2} )]$

                                  $= \lim_{x \to y} \frac{2}{2(x-y)}[ 2cos(\frac{x+y}{2} )sin(\frac{x-y}{2} )]\\\\=\lim_{x \to y} \frac{1}{(\frac{x-y}{2}) }[ cos(\frac{x+y}{2} )sin(\frac{x-y}{2} )]$

                                  $=\lim_{x \to y} [ \frac{sin(\frac{x-y}{2} )}{\frac{x-y}{2} }]\lim_{x \to y} [ cos(\frac{x+y}{2} )]\\\\=\lim_{(x-y) \to 0} [\frac{sin(\frac{x-y}{2} )}{\frac{x-y}{2} }]\lim_{x \to y} [ cos(\frac{x+y}{2} )]$

Assume:   $\frac{(x-y)}{2} =t$   because   ${(x-y) \to 0}$

→Therefore by Using the formula:  $\lim_{t \to 0} \frac{sint}{t}=1$

                            $\therefore Limit=(1)(cos\frac{2y}{2} )$    

                            $\therefore Limit=cosy$

Therefore the value of the limit is equal to cos(y).

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