Find the value of \[\underset{x\to -\infty }{\mathop{\lim }}\,\left[ \dfrac{{{x}^{4}}\sin \left( \dfrac{1}{x} \right)+{{x}^{2}}}{\left( 1+{{\left| x \right|}^{3}} \right)} \right]=?\]

Hint: The given problem is related to evaluation of limit of a function. Try to eliminate the trigonometric terms using formulae of limits for trigonometric functions, like $\underset{x\to 0}{\mathop{\lim }}\,\dfrac{\sin x}{x}=1$ . Then, take the variable with highest power common in the numerator and the denominator. Then substitute the limits.

Complete step-by-step answer:
The given limit to be evaluated is \[\underset{x\to -\infty }{\mathop{\lim }}\,\left[ \dfrac{{{x}^{4}}\sin \left( \dfrac{1}{x} \right)+{{x}^{2}}}{\left( 1+{{\left| x \right|}^{3}} \right)} \right]\] . We know, when x tends to $-\infty $ , $\dfrac{1}{x}$ tends to 0. We also know, we can write ${{x}^{4}}$ as $\dfrac{{{x}^{3}}}{\dfrac{1}{x}}$ ………… (i) and $\underset{x\to 0}{\mathop{\lim }}\,\dfrac{\sin x}{x}=1$……….. (ii) . Now, let the value of the limit be L. So, $L=\underset{x\to -\infty }{\mathop{\lim }}\,\left[ \dfrac{{{x}^{4}}\sin \left( \dfrac{1}{x} \right)+{{x}^{2}}}{\left( 1+{{\left| x \right|}^{3}} \right)} \right]$ . Now, from (i), we can write $L=\underset{x\to -\infty }{\mathop{\lim }}\,\left[ \dfrac{\dfrac{{{x}^{3}}}{\dfrac{1}{x}}\sin \left( \dfrac{1}{x} \right)+{{x}^{2}}}{\left( 1+{{\left| x \right|}^{3}} \right)} \right]$
$\Rightarrow L=\underset{x\to -\infty }{\mathop{\lim }}\,\left[ \dfrac{{{x}^{3}}\dfrac{\sin \left( \dfrac{1}{x} \right)}{\dfrac{1}{x}}+{{x}^{2}}}{\left( 1+{{\left| x \right|}^{3}} \right)} \right]$
Now, from (ii), we know $\underset{x\to 0}{\mathop{\lim }}\,\dfrac{\sin x}{x}=1$ . So, we can write $L=\underset{x\to -\infty }{\mathop{\lim }}\,\left[ \dfrac{{{x}^{3}}+{{x}^{2}}}{\left( 1+{{\left| x \right|}^{3}} \right)} \right]$ .
Now, as \[x\to -\infty\] , the value of x is negative. So, by the definition of modulus function, we can say \[\left| x \right|=-x\].
\[\Rightarrow {{\left| x \right|}^{3}}=-{{x}^{3}}\]
\[\Rightarrow 1+{{\left| x \right|}^{3}}=1-{{x}^{3}}\]
We will substitute the value of \[1+{{\left| x \right|}^{3}}\] in the denominator. So, we get the limit as $L=\underset{x\to -\infty }{\mathop{\lim }}\,\left[ \dfrac{{{x}^{3}}+{{x}^{2}}}{\left( 1-{{x}^{3}} \right)} \right]$ . Now, the function consists of polynomials. So, we will take the variable with the highest power, i.e.${{x}^{3}}$ , common from the numerator as well as the denominator. So, we get:
$L=\underset{x\to -\infty }{\mathop{\lim }}\,\left[ \dfrac{{{x}^{3}}\left( 1+\dfrac{1}{x} \right)}{{{x}^{3}}\left( \dfrac{1}{{{x}^{3}}}-1 \right)} \right]$
$\Rightarrow L=\underset{x\to -\infty }{\mathop{\lim }}\,\left[ \dfrac{\left( 1+\dfrac{1}{x} \right)}{\left( \dfrac{1}{{{x}^{3}}}-1 \right)} \right]$
Substituting $x=-\infty $ in the limit, we get :
$L=\dfrac{1+\dfrac{1}{-\infty }}{\dfrac{1}{-{{\infty }^{3}}}-1}$
$\Rightarrow L=\dfrac{1+0}{0-1}$
$\Rightarrow L=-1$
Hence, the value of the limit is equal to -1.

Note: While solving problems related to limits, try to find out the terms in the function, which can lead to attainment of indeterminate forms and then try to eliminate those terms. Also, while performing simplification or substitution, always take care of sign, Sign mistakes are very common and can lead to incorrect answers.