Question

$\large \boxed{ \boxed{ \mathfrak{solve \: \: the \: \: given \: \: question}}}$

$\lim_{x \to 0} \dfrac{ \sin(2x) + sin(x)}{ { \sin {}^{3} (x) }^{} }$




Solution required ~ ​

Answer 1

Answer:

$\lim_{x \to 0} \frac{2\cos(x)+1}{\sin^2 (x) } = \infty$

and the limit does not exist

Step-by-step explanation:

We are given the following limit

$\lim_{x \to 0} \frac{\sin(2x)+\sin(x)}{\sin^3 (x)}$

We can't just merely consider

$\lim_{x \to 0} f(x)=g(x)$

and then solve

$\lim_{x \to 0} \frac{\sin(2x)+\sin(x)}{\sin^3 (x)} = \frac{\sin(0)+\sin(0)}{\sin^3 (0)} =\frac{0}{0} =0$

This is an indeterminate form. Wrong.

=========================================

We have the identity

$\sin(2x) = 2\sin(x)\cos(x)$, therefore we can substitute it in the limit

$\lim_{x \to 0} \frac{\sin(2x)+\sin(x)}{\sin^3 (x)} = \lim_{x \to 0} \frac{2\sin(x)\cos(x)+\sin(x)}{\sin^3 (x)}$

then

$\lim_{x \to 0} \frac{2\sin(x)\cos(x)+\sin(x)}{\sin^3 (x)} = \lim_{x \to 0} \frac{\sin(x)(2\cos(x)+1)}{\sin^3 (x)} = \boxed{ \lim_{x \to 0} \frac{2\cos(x)+1}{\sin^2 (x)} }$

Now, differently from the first case,

we have $2\cos(x)+1 \longrightarrow 3$ as $x \longrightarrow 0$ because $\cos(0)=1$

On the other hand,  $\sin^2(x)\longrightarrow 0$ as $x \longrightarrow 0$ because $\sin(0)=0$

In fact,

$\lim_{x \to 0} \frac{2\cos(x)+1}{\sin^2 (x) } = \lim_{x \to 0} \frac{1}{\sin^2 (x) }\cdot (2\cos(x)+1) = \infty \cdot 3 = \infty$

Answer 2

>> Answer

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$\:$

Answer in the picture.