Question

Find the value of m so that

lim (sin 2x + m sin 3x)/ x^3

x→0

is finite.

Answer 1

$\begin{aligned} &\operatorname{lim}_{x \rightarrow 0} \frac{\sin 2 x+m \sin 3x}{x^{3}} \\ &lim_{x \rightarrow 0} \frac{[2 \sin x\cos x+m(3\sin x-4\sin^3x)]}{x^{3}} \\ &lim_{x \rightarrow 0} \frac{\sin x[2 \cos x+m(3-4\sin^2x)]}{x^{3}} \\ &lim_{x \rightarrow 0} \frac{[2 \cos x+m(3-4\sin^2x)]}{x^{2}} \\ &\{\because lim_{x \rightarrow 0} \frac{\sin x}{x}=1 \}\\ \end{aligned}$

As $x \rightarrow 0$ , denominator tends to , so the numerator also tends to 0.

Thus, $lim_{x \rightarrow 0} \ [2 \cos x+m(3-4\sin^2x)]=0\\ 2+3m=0\\ m=\frac{-2}{3}$