lim x → − ∞ { x 4 sin ( 1 x ) + x 2 1 + | x | 3 } is equal to Options: A . &nbsp 2 B . &nbsp 1 C . &nbsp − 1 D . &nbsp does not exist

Answer: Option C : C lim x → − ∞ [ x 4 sin ( 1 x ) + x 2 1 + | x | 3 ] lim x → − ∞ x 3 1 + | x | 3 [ x sin ( 1 x ) + 1 x ] As x → − ∞ x 3 1 + | x | 3 = − 1 x s i n ( 1 x ) + 1 x = 1 ⇒ l i m x → − ∞ x 3 1 + | x | 3 ( x s i n 1 x + 1 x ) = − 1

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

Discussion Forum : Limits Continuity And Differentiability

Question -

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

Options:

A .

2

B .

1

C .

- 1

D . does not exist

Answer: Option C
:
C

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}]

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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Discussion Forum : Limits Continuity And Differentiability

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