Limits, Continuity and Differentiability
79543
The value of \(\lim _{x \rightarrow 0}(\cos x+a \operatorname{sinbx})^{1 / x}\) is
1 1
2 \(\mathrm{ab}\)
3 \(\mathrm{e}^{\mathrm{ab}}\)
4 \(\mathrm{e}^{\mathrm{b} / \mathrm{a}}\)
Explanation:
(C) : Given,
\(\lim _{x \rightarrow 0}(\cos x+a \sin b x)^{1 / x}\)
If \(\lim _{x \rightarrow a} f(x)=\lim _{x \rightarrow a} g(x)=0\) such that \(\lim _{x \rightarrow a} \frac{f(x)}{g(x)}\) exists.
\(\left[\therefore \lim _{\mathrm{x} \rightarrow \mathrm{a}}[1+\mathrm{f}(\mathrm{x})]^{\frac{1}{\mathrm{~g}(\mathrm{x})}}=\mathrm{e}^{\lim _{\mathrm{x \rightarrow a}} \frac{\mathrm{f}(\mathrm{x})}{\mathrm{g}(\mathrm{x})}}\right]\)
Let us add and subtract 1 to the given expression.
\(=\lim _{x \rightarrow 0}(1+\cos x+a \sin b x-1)^{1 / x}\)
Here, \(f(x)=\cos x+a \sin b x-1\)
and \(g(x)=x\)
\(=\mathrm{e}^{\lim _{x \rightarrow 0}\left[\frac{\cos x+a \sin b x-1}{x}\right]}=\mathrm{e}^{\lim _{x \rightarrow 0}\left[\frac{a \sin b x}{x}-\left(\frac{1-\cos x}{x}\right)\right]}\)
\(=\mathrm{e}^{\lim _{x \rightarrow 0}\left[\frac{a b \sin b x}{b x}-\frac{2 \sin ^{2} x / 2}{x}\right]}=\mathrm{e}^{\lim _{x \rightarrow 0}\left[\frac{a b \sin b x}{b x}-0\right]}=e^{a b}\)
