Chapter 2, Problem 6RE

To determine

To find: The value of given limit.

Answer to Problem 6RE

The value of $\underset{x\to \pm \infty }{\lim }\frac{2x^2+3}{5x^2+7}$ is $\frac{2}{5}$ .

Explanation of Solution

Given information: The given expression is $\underset{x\to \pm \infty }{\lim }\frac{2x^2+3}{5x^2+7}$ .

Calculation:

Eliminate the integers in the expression as $x$ tends to infinity.

  $\underset{x\to \pm \infty }{\lim }\frac{2x^2+3}{5x^2+7}=\underset{x\to \pm \infty }{\lim }\frac{2x^2}{5x^2}$

Further solve.

  $\underset{x\to \pm \infty }{\lim }\frac{2x^2}{5x^2}=\frac{2}{5}$

Therefore, the value of $\underset{x\to \pm \infty }{\lim }\frac{2x^2+3}{5x^2+7}$ is $\frac{2}{5}$ .