X - -limx4 sin 1-x -x2-1-x-3 is equal to

X → ∞lim [ x^4 sin ( 1/x )+x^2/1+|x|^3 ] x → ∞limx^3/1+|x|^3 [ x sin ( 1/x ) +1/x ] As x → ∞ x^3/1+|x|^3= 1 x sin ( 1/x ) + 1/x=1 ⇒x → ∞limx^3/1+|x|^3 ...

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Standard XIII

Mathematics

Standard Limits to Remove Indeterminate Form

x → -∞limx4 s...

Question

$lim x \to - \infty {\frac{x^{4} sin (\frac{1}{x}) + x^{2}}{1 + | x |^{3}}}$ is equal to

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Solution

The correct option is C $- 1$
$lim x \to - \infty [\frac{x^{4} sin (\frac{1}{x}) + x^{2}}{1 + | x |^{3}}]$

$lim x \to - \infty \frac{x^{3}}{1 + | x |^{3}} [x sin (\frac{1}{x}) + \frac{1}{x}]$

As $x \to - \infty$

$\frac{x^{3}}{1 + | x |^{3}} = - 1$

$x s i n (\frac{1}{x}) + \frac{1}{x} = 1$

$\Rightarrow l i m x \to - \infty \frac{x^{3}}{1 + | x |^{3}} (x s i n \frac{1}{x} + \frac{1}{x}) = - 1$

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limx→−∞{x4sin(1x)+x21+|x|3} is equal to

The correct option is C −1limx→−∞[x4sin(1x)+x21+|x|3] limx→−∞x31+|x|3[xsin(1x)+1x] As x→−∞ x31+|x|3=−1 x sin(1x)+1x=1 ⇒limx→−∞x31+|x|3(x sin1x+1x)=−1

$lim x \to - \infty {\frac{x^{4} sin (\frac{1}{x}) + x^{2}}{1 + | x |^{3}}}$ is equal to