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2015-09-14T18:39:25+05:30

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Lim √x2 + x - √x 
x->∞
let x=1/y so that y->0, when x->

Lim √1/y2 + 1/y - √1/y 
y->0
Lim 1/y + 1/y - √1/y 
y->0
Lim (1+1)/y - √1/y 
y->0
Lim 2/y - √1/y 
y->0
Lim 2/y - 1/√y 
y->0
Lim (2√y - y)/y√y 
y->0
Lim (2√y - y)/y√y x (2√y + y)/(2√y + y)
y->0
Lim (4y - y2)/y√y(2√y + y) 
y->0
Lim (4y - y2)/(2y2 + y2√y) 
y->0
dividing each term with the highest power of y
Lim (4y/y2 - y2/y2)/(2y2/y2 + y2√y/y2) 
y->0
Lim (4/y - 1)/(2 + √y) 
y->0
(4/0 - 1)/(2 + 0)
0 - 1 / 2
-1/2 it is the answer....
1 5 1
4/y -> infinity and not 0...
no no
oops haa
yes ri8
thanks
2015-09-15T02:38:08+05:30

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Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.
The given question is not clearly specified.

 \lim_{x \to \infty} \sqrt{x^2+x} -\sqrt{x}\\\\= \lim_{x \to \infty}\ \ x*[\sqrt{1+\frac{1}{x}}-\frac{1}{\sqrt{x}}}]\\\\= \lim_{x \to \infty}  x * [\sqrt{1+0}-0]= \lim_{x \to \infty} x\\\\=\infty
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\sqrt{x^2}+x-\sqrt{x}=x+x-\sqrt{x}=2x-\sqrt{x}=x*(2-\frac{1}{\sqrt{x}})\\\\ \lim_{x \to \infty} x*(2-\frac{1}{\sqrt{x}} )\\\\=\lim_{x \to \infty} x (2-0)=\lim_{x \to \infty} x*2\\\\=\infty



1 5 1
*under root x2+x
and + x will be out of under root
use parentheses ( ) , [ ] { } brackets to clearly write the expression please
Lim √x2 + x - √x
x->∞
this is the question