Question

The value of $\underset{x\to -\infty }{lim}\sqrt{25x^2-3x}+5x$ is

• A. $\frac{3}{5}$
• B. $\frac{3}{10}$
• C. $-\frac{3}{10}$
• D. does not exist

Answer 1

The correct option is B $\frac{3}{10}$

$L=\underset{x\to -\infty }{lim}\sqrt{25x^2-3x}+5x,(\infty -\infty \text{form})=\underset{x\to -\infty }{lim}\frac{25x^2-3x-25x^2}{\sqrt{25x^2-3x}-5x}=\underset{x\to -\infty }{lim}\frac{-3x}{-x\left(\sqrt{25-\frac{3}{x}}\right)-5x}(\because x\to -\infty \Rightarrow |x|=-x)=\frac{3}{\sqrt{25-0}+5}=\frac{3}{10}$