Question

Hello find the limits.
Better to be answered by one who know!
Thanks

Answer 1

Good Night!

lim Sin ( a + 2x ) - 2 Sin ( a + x ) + Sin a

x->0 ----------------------------------------

x²

Put x = 0 We get 0/0 Form.

Now, use L'HOSPITALS RULE Which states that Differentiate Numerator and Denominator separately till we come out from 0/0 form.

lim 2 Cos ( a + 2x ) - 2 Cos ( a + x )

x->0 --------------------------------------

2x

lim Cos ( a + 2x ) - Cos ( a + x )

x->0 -----------------------------------

x

If we Again put x = 0 we get Again 0/0 form. Now we use Again L'Hospitals rule.

lim -2 Sin ( a + 2x ) + Sin ( a + x )

x->0 ----------------------------------

1

= -2 Sin ( a + 0 ) + Sin ( a )

= -2 Sin ( a ) + Sin ( a )

= - Sin ( a )

Answer 2

Step-by-step explanation:

Important Formulas:

(i) sin(a + b) = sina cosb + cosa sinb

(ii) cos 2a = cos²a - sin²a

Solution:

$Given: \frac{sin(a+2x) - 2sin(a + x) - sina}{x^2}$

$\Longrightarrow \frac{sina\cos2x + sin2x\ cosa - (2sina \ cosx + sin x \ cosa) + sina }{x^2}$

$\Longrightarrow \frac{sina(cos^2x - sin^2x) + 2sinxcosa - 2sinacosx - 2sinxcosa+sina}{x^2}$

$\Longrightarrow \frac{sina(1 - 2sin^2x-2cosx+1)+cosa(2sinxcosx-2sinx)}{x^2}$

$\Longrightarrow \frac{sina(2-2sin^2x-2cosx) +cosa(2sinxcosx-2sinx)}{x^2}$

$\Longrightarrow 2sina[(\frac{1-cosx}{x^2})-\frac{sin^2x}{x^2}] + 2cosa[\frac{sinx}{x}\frac{cosx-1}{x}]$

$\Longrightarrow 2sina[\frac{cosx}{2}-cos2x] + 2cosa[\frac{sinx}{x}\frac{cosx-1}{x}]$

Applying Limits:

$\lim_{x \to 0} \ 2sina[\frac{cosx}{2}-cos2x] +2cosa[\frac{sinx}{x}\frac{cosx-1}{x}]$

$\Longrightarrow 2sina(\frac{1}{2} - 1) + 0$

$\Longrightarrow 2sina(\frac{-1}{2})$

$\Longrightarrow \textbf {-sina}$

Hope it helps!