Question

$\underset{x\to -\infty }{\lim }\left[\frac{\left(2x-1\right)}{\left(\sqrt{x^2+2x+1}\right)}\right]=$

• A. $2$
• B. $-2$
• C. $1$
• D. $-1$
• E. $0$

Answer 1

The correct option is B

$-2$

Explanation for the correct option:

Find the value of $\underset{x\to -\infty }{\lim }\left[\frac{\left(2x-1\right)}{\left(\sqrt{x^2+2x+1}\right)}\right]$

Consider the given Equation as

$I=\underset{x\to -\infty }{\lim }\left[\frac{\left(2x-1\right)}{\left(\sqrt{x^2+2x+1}\right)}\right]$

Put $x=-t$, then the above Equation becomes,

$$\begin{gathered} =\underset{t\to \infty }{\lim }\left[\frac{\left(2\left(-t\right)-1\right)}{\left(\sqrt{{\left(-t\right)}^2+2\left(-t\right)+1}\right)}\right] \\ =\underset{t\to \infty }{\lim }\left[\frac{\left(-2t-1\right)}{\left(\sqrt{t^2-2t+1}\right)}\right] \\ =\underset{t\to \infty }{\lim }\left[\frac{t\left(-2-{\displaystyle \frac{1}{t}}\right)}{t\left(\sqrt{1-{\displaystyle \frac{2}{t}}+{\displaystyle \frac{1}{t^2}}}\right)}\right] \\ =\underset{t\to \infty }{\lim }\left[\frac{\left(-2-{\displaystyle \frac{1}{t}}\right)}{\left(\sqrt{1-{\displaystyle \frac{2}{t}}+{\displaystyle \frac{1}{t^2}}}\right)}\right] \end{gathered}$$

$$\begin{gathered} \text{At}t\to \infty \\ \frac{1}{t}\to \frac{1}{\infty }\to 0 \end{gathered}$$

Then,

$$\begin{gathered} I=\left[\frac{\left(-2-{\displaystyle 0}\right)}{\left(\sqrt{1-{\displaystyle 0}+{\displaystyle 0}}\right)}\right] \\ =\left[\frac{-2}{1}\right] \\ =-2 \end{gathered}$$

Hence, the correct answer is option B.