Question

$\underset{x\to 0}{lim}\frac{3\sqrt{1+x^2}-4\sqrt{1-2x}}{x+x^2}$ is equal to

• A. $\frac{1}{4}$
• B. $\frac{1}{2}$
• C. $1$
• D. $Doesnotexists$

Answer 1

The correct option is D $Doesnotexists$

$\underset{0\to 0}{lim}\frac{3\sqrt{1+x^2}-4\sqrt{1-2x}}{x(x+1)}=\underset{x\to 0}{lim}\frac{9(1+x^2)-16(1-2x)}{x(x+1)(3\sqrt{1+x^2}+4\sqrt{1-2x})}$ (Rationalising)

$=\underset{x\to 0}{lim}\frac{9x^2+32x-7}{x(x+1)(3\sqrt{1+x^2}+4\sqrt{1-2x})}$

$=\frac{-7}{0\left(1\right)\left(7\right)}=\infty$ (on puttinf $x=0$)

$\therefore$ Option $D$ is correct