Question

please solve limits sum​

Answer 1

Step-by-step explanation:

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Answer 2

Answer:

The solved limit sum is $lim_{x\longrightarrow{0}}\frac{sin(a+x)+sin(a-x)-2sina}{xsinx}=-sina$

Step-by-step explanation:

Given limit sum is $lim_{x\overrightarrow{0}}\frac{sin(a+x)+sin(a-x)-2sina}{xsinx}$

To solve the given limit sum :

Now taking $sin(a+x)+sin(a-x)-2sina$

$=2sinacosa-2sina$ ( by using the formula $sin(x+y)+sin(x-y)=2sinxcosy$ here x=a and y=x )

$=2sina(cosx-1)$

$=-2sina(-cosx-(-1))$

$=-2sina(1-cosx)$

Multiply and divide by conjugate we get

$=\frac{-2sina(1-cosx)\times (1+cosx)}{1+cosx}$

$=\frac{-2sina(1^2-cos^2x)}{1+cosx}$  ( by using the formula $(a-)(a+b)=a^2-b^2$ )

$=\frac{-2sina(1-cos^2x)}{1+cosx}$  

$sin(a+x)+sin(a-x)-2sina=\frac{-2sinasin^2x}{1+cosx}$ ( by using the formula $sin^2x=1-cos^2x$ )

Now taking $lim_{x\longrightarrow{0}}$ on both sides an d divided by xsinx on both sides we get

$lim_{x\longrightarrow{0}}\frac{sin(a+x)+sin(a-x)-2sina}{xsinx}=lim_{x\longrightarrow{0}}\frac{-2sinasin^2x}{(1+cosx)(xsinx)}$

$=lim_{x\longrightarrow{0}}\frac{-2sinasinx}{(1+cosx)(x)}$

$=lim_{x\longrightarrow{0}}\frac{-2sina}{(1+cosx)}\times lim_{x\longrightarrow{0}}\frac{sinx}x}$

 ( here using $lim_{x\longrightarrow{0}}\frac{sinx}{x}=1$ )

$=\frac{-2sina}{(1+1)}$ ( by cos0=1 )

$=-\frac{2sina}{2}$

$=-sina$

Therefore $lim_{x\longrightarrow{0}}\frac{sin(a+x)+sin(a-x)-2sina}{xsinx}=-sina$