Question

limits x tends to infinite under root x square + x -under  root x.
plss urgent ..

Answer 1

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....

Answer 2

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