Question

$\underset{x\to \infty }{lim}(\sqrt{x^2+x}-x)$ equals

• A.
• B.
• C.
• D.

Answer 1

The correct option is B

$\underset{x\to \infty }{lim}(\sqrt{(x^2+x)}-x)$

Put x = $\frac{1}{t}$

$\therefore$ = $\underset{t\to 0}{lim}(\sqrt{(\frac{1}{t^2}+\frac{1}{t})}-\frac{1}{t})$
On rationalisation we get the above expression = $\frac{1}{2}$

Now, Let's Check the options -

Option (a):

$\underset{x\to 0}{lim}\frac{x+In(1-x)}{x^2}$

= $\underset{x\to 0}{lim}\frac{1-\frac{1}{(1-x)}}{2x}$ (by L'Hospital's rule)

=$-\frac{1}{2}$

Option (b):

= $\underset{x\to 0}{lim}\frac{e^{-x}-1+x}{x^2}$

= $\underset{x\to 0}{lim}\frac{-e^{-x}-0+1}{2x}$ (by L'Hospital's rule)

= $\underset{x\to 0}{lim}\frac{e^{-x}+0}{2}$ = $\frac{1}{2}$

Option (c):

= $\underset{x\to 0}{lim}\frac{-\sqrt{x}}{\sqrt{x}+\sqrt{(x^2+2x)}}$

= $\underset{x\to 0}{lim}\frac{1}{1+\sqrt{(x+2)}}$ = -$\frac{1}{(1+\sqrt{2})}$

Option (d):

= $\underset{x\to 0}{lim}\frac{cos{x}^2-1}{x^4}$

Put $x^2$ = $t$

$\underset{t\to 0}{lim}\frac{cost-1}{t^2}$

= $\underset{t\to 0}{lim}-\frac{sint}{2t}$ (by L'Hospital's rule)

= $-\frac{1}{2}$