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