Evaluate the given limit  $\lim\limits_{x \to 0} \large\frac{ \sin \: ax+bx}{ax+ \sin \: bx}$$a,b, a+b \neq0 \begin{array}{1 1}1 \\ 0 \\ b \\ a \end{array} 1 Answer At x = 0 the value of the given function is \large\frac{0}{0}. \lim\limits_{x \to 0} \large\frac{ \sin \: ax+bx}{ax+ \sin \: bx}$$= \lim\limits_{x \to 0} \large\frac{ \bigg(\Large\frac{ \sin \: ax}{ax}\bigg)\normalsize ax+bx}{ax+ bx \bigg( \Large\frac{ \sin \: bx}{bx}\bigg)}$
As $\: x \to 0 \Rightarrow ax \to 0 \: and \: bx \to 0$
$= \large\frac{ \bigg( \lim\limits_{ax \to 0} \Large\frac{\sin \: ax}{ax} \bigg) \normalsize\; \times\; \lim\limits_{x \to 0} (ax) + \lim\limits_{x \to 0} bx}{ \lim\limits_{x \to 0}ax+\lim\limits_{x \to 0}bx \bigg( \lim\limits_{bx \to 0} \Large\frac{\sin \: bx}{bx} \bigg)}$
$= \large\frac{ \lim\limits_{x \to 0}(ax)+ \lim\limits_{x \to 0}bx}{\lim\limits_{x \to 0}ax+\lim\limits_{x \to 0}bx}$$\quad \quad \bigg[ \lim\limits_{x \to 0} \large\frac{\sin \: x}{x}$$ = 1\bigg]$
$= \large\frac{\lim\limits_{x \to 0}(ax+bx)}{\lim\limits_{x \to 0}(ax+bx)}$
$\lim\limits_{x \to 0}(1) = 1$
edited May 16, 2014