Question

The limit of $\{\frac{1}{x}\sqrt{1-x}-\sqrt{1+\frac{1}{x^2}}\}$ as $x\to 0$

• A. Does not exist
• B. Is equal to $\frac{1}{2}$
• C. Is equal to 0
• D. Is equal to 1

Answer 1

The correct option is A Does not exist

Let $p\left(x\right)=\frac{1}{x}\sqrt{1-x},q\left(x\right)=\sqrt{1+\frac{1}{x^2}}$

${lim}_{x\to 0^{+}}p\left(x\right)={lim}_{x\to 0^{+}}\frac{1}{x}\sqrt{1-x}=\infty \times \sqrt{1-10}$

$=\infty$

${lim}_{x\to 0^{-}}p\left(x\right)=-\infty \times \sqrt{1-0}=-\infty$

as $x\to 0^{+}\frac{1}{x}\to +\infty$

$x\to 0^{-}\frac{1}{x}\to -\infty$

$\Rightarrow p\left(x\right)$ is discontinuous as $x\to 0$

Let $f\left(x\right)=p\left(x\right)-q\left(x\right)$

Since $p\left(x\right)$ is discontinuous $f\left(x\right)$ is also discontinuous by Algebra of continuous functions.