Q. 18 - Calculus

Question

$\infty -\infty$that approaches (a) $0$ (b) $5,$ (c) $\infty .$

Step-by-Step Solution

Answer

$\infty -\infty$that approaches to (c) $\infty .$

Step 1. Given information

The given expression is $\infty -\infty ,$ we need to write $\infty -\infty$that approaches to (a) $0$ (b) $5$ (c) $\infty .$

Step 2. The explanation for the correct option

Take an example to check the limits of the given form.

$\begin{array}{rcl}\underset{x\to \infty }{\lim }\frac{5\left(x^2 - 1\right)}{x^2 - x}& =& \underset{x\to \infty }{\lim }\frac{5{\displaystyle \frac{d}{dx}}\left(x^2 - 1\right)}{\frac{d}{dx}\left(x^2 - x\right)} \left[numartor is form of \infty -\infty \right]\\ & =& \underset{x\to \infty }{\lim }\frac{5\left(2x\right)}{\left(2x -1\right)}\\ & =& \underset{x\to \infty }{\lim }\frac{5\left(2x\right)}{x\left(2 -\frac{1}{x}\right)}\\ & =& \underset{x\to \infty }{\lim }\frac{10}{\left(2 -\frac{1}{x}\right)}\\ & =& \frac{10}{2}\\ & =& 5\end{array}$

And,

Like $\underset{x\to \infty }{\lim }\left[\sqrt{x^2-x}-x\right].$

        $=\underset{x\to \infty }{\lim }\left[\sqrt{x^2}\sqrt{1-\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$x$}\right.}-x\right].$

         $=\underset{x\to \infty }{\lim }x\left[\sqrt{1-\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$x$}\right.}-1\right].$

          $=\underset{x\to \infty }{\lim }\frac{\sqrt{1-{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$x$}\right.}}-1}{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$x$}\right.}}.$

          $=\underset{x\to \infty }{\lim }\frac{\left({\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\sqrt{1-{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$x$}\right.}}\right)\left(+{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$x^2$}\right.}\right)}{-{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$x^2$}\right.}}.$[Applying L Hospital rule].

          $=-\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right..$