Question

Evaluate $\displaystyle \bf \lim_{ \bf x \to 0} \bf \dfrac{\sin ax + bx}{\sin bx + ax}$ ​

Answer 1

$\huge\bold{\color{aqua}{Solution :-}}$

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

$\bold{\purple{\text{Dividing numerator and denominator by x}}}$

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

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

$\Rightarrow\bold{\lim_{x\to 0}\frac{a\bigg(\frac{sin\:ax}{ax}\bigg)\:+\:b}{b\bigg(\frac{sin\:bx}{bx}\bigg)\:+\:a}}$

$\Rightarrow\bold{\frac{a\bigg(\lim_{x\to 0}\frac{sin\:ax}{ax}\bigg)\:+\:b}{b\bigg(\lim_{x\to 0}\frac{sin\:bx}{bx}\bigg)\:+\:a}}$

$\Rightarrow\bold{\frac{a(1)+b}{b(1)+a}\qquad\qquad\therefore\orange{\small{\lim_{A\to 0}\frac{sin\:A}{A}=1}}}$

$\Rightarrow\bold{\frac{a+b}{a+b}}$

$\Rightarrow\bold{\frac{\red{\cancel{\blue{a+b}}}^{\green{\:\:1}}}{\red{\cancel{\blue{a+b}}}^{\green{\:\:1}}}}$

$\Rightarrow\bold{\frac{1}{1}}$

$\Rightarrow\bold{\green{\boxed{\pink{1}}}}$

Answer 2

Question:-

$\sf \: Evaluate: \sf \lim_{ x \to 0} \sf\dfrac{sin \: ax + bx}{ sin \: bx + ax}$

Given:-

$\sf \lim_{ x \to 0} \sf\dfrac{\sin ax + bx}{\sin bx + ax}$

Solution:-

We have:

$\sf \lim_{ x \to 0} \sf\dfrac{sin \: ax + bx}{sin \: bx + ax}$

Dividing numerator and denominator by x, we get;

$\sf \large\lim_{ x \to 0} \: \frac{ \frac{sin \: ax}{x} + b }{a + \frac{sin \: bx}{x} } \sf \large = \lim_{ x \to 0} \: \frac{ \frac{sin \: ax}{ax} \: \times a + b}{a + ( \frac{sin \: bx}{bx} \times b )}$

$\sf \large = \frac{a (\lim_{ x \to 0} \: \frac{sin \: ax}{ax}) + b }{a + b(\lim_{ x \to 0} \: \frac{sin \: bx}{bx} )}$

$\sf \large = \frac{a(1) + b}{a + b(1)}$

$\sf \large = {{ \cancel{ \frac{a + b}{a + b}}}}$

$\sf \large = 1$

Answer:-

$\sf \: \red{Hence, \sf \lim_{ x \to 0} \sf\dfrac{\sin ax + bx}{\sin bx + ax} = 1.}$

Hope you have satisfied. ⚘