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Home  >>  JEEMAIN and NEET  >>  Mathematics  >>  Class11  >>  Limits and Derivatives

$\lim\limits_{x\to -\infty}\bigg[\large\frac{x^4\sin(1/x)+x^2}{(1+|x|^3)}\bigg]$=

$(a)\;-1\qquad(b)\;1\qquad(c)\;0\qquad(d)\;2$

1 Answer

$\lim\limits_{x\to -\infty}\bigg[\large\frac{x^4\sin(1/x)+x^2}{1+|x|^3}\bigg]$
Let $L=\lim\limits_{x\to -\infty}\large\frac{x^3}{1+|x|^3}$$[x\sin(1/x)+1/x]$
$\Rightarrow \lim\limits_{x\to -\infty}\large\frac{x^3}{|xx|^3}\bigg[\large\frac{1}{1+\Large\frac{1}{|x|x^2}}\bigg]$$[x\sin(1/x)+1/x)\bigg]$
$\Rightarrow \lim\limits_{x\to -\infty}\large\frac{x^3}{|x|^3}$$.1=\lim\limits_{x\to -\infty}\large\frac{x^3}{-x^3}$
$\Rightarrow -1$
Hence (a) is the correct answer.
answered Dec 20, 2013 by sreemathi.v
 

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