Question

Evaluate the Given limit: $\lim_{x\rightarrow 0}$ sinax =$\frac{bx}{ax}$+ sinbx a,b,a+b ≠ 0

Answer 1

$\lim_{x\rightarrow 0}\,sin\,ax$ = $\frac{bx}{ax}+sin\,bx$
At x = 0, the value of the given function takes the form 0/0.
Now,
= $\lim_{x\rightarrow 0}$ $sin\,ax$ = $\frac{bx}{ax}+sin\,bx$
= $\lim_{x\rightarrow 0}$ ($\frac{sin\,ax}{ax}$) ax + $\frac{bx}{ax}$ + bx($\frac{sin\,bx}{bx}$)
=($\lim_{ax\rightarrow 0}$ $\frac{sin\,ax}{ax}$) $\times$ $\lim_{x\rightarrow 0}$(ax) + $\lim_{x\rightarrow 0}$bx / $\lim_{x\rightarrow 0}$(ax) + ($\lim_{bx\rightarrow 0}$ $\frac{sin\,bx}{bx}$) [As x$\rightarrow$0$\Rightarrow$ax$\rightarrow$0 and bx$\rightarrow$0]
=$\frac{\lim_{x\rightarrow 0}(ax)\lim_{x\rightarrow 0}bx}{\lim_{x\rightarrow 0}ax+\lim_{x\rightarrow 0}bx}$
=$\frac{\lim_{x\rightarrow 0}(ax+bx)}{\lim_{x\rightarrow 0}(ax+bx)}$
=$\lim_{x\rightarrow 0}$ (1)
=1