Question

please help me friends​

Answer 1

SOLUTION

TO DETERMINE

$\displaystyle \lim_{x \to 0} \: \frac{ \sin ax + bx}{ax + \sin bx}$

EVALUATION

$\displaystyle \lim_{x \to 0} \: \frac{ \sin ax + bx}{ax + \sin bx}$

$= \displaystyle \lim_{x \to 0} \: \frac{ \frac{ \sin ax\: }{x} + b}{a+ \frac{ \sin bx\: }{x} }$

( Dividing numerator and denominator both by x )

$= \displaystyle \lim_{x \to 0} \: \frac{ a \: .\frac{ \sin ax\: }{ax} + b}{a+ b \: . \: \frac{ \sin bx\: }{bx} }$

$= \displaystyle \frac{ a \:\displaystyle \lim_{x \to 0} \: \frac{ \sin ax\: }{ax} + b}{a+ b \: \displaystyle \lim_{x \to 0} \: \frac{ \sin bx\: }{bx} }$

Let u = ax and v = bx

Then as x → 0 we have u → 0 and v → 0

Therefore

$\displaystyle \lim_{x \to 0} \: \frac{ \sin ax + bx}{ax + \sin bx}$

$= \displaystyle \frac{ a \: . \:\displaystyle \lim_{u \to 0} \: \frac{ \sin u\: }{u} + b}{a+ b \: .\: \displaystyle \lim_{v \to 0} \: \frac{ \sin v\: }{v} }$

$= \displaystyle \frac{(a \times 1) +b }{a + (b \times 1)} \: \: \: \bigg( \because \: \displaystyle \lim_{x \to 0} \: \frac{ \sin x}{x} = 1 \bigg)$

$= \displaystyle \frac{a + b}{a + b}$

$= 1$

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