Question

# Approximately, what is $\mathop {\lim }\limits_{x \to \infty } \dfrac{{\sqrt 7 {x^2} + 3x - 2}}{{{x^2} + 5}}$?A. 2.32B. 2.43C. 2.54D. 2.65E. 2.76

Hint: In this problem, take ${x^2}$ common from numerator and denominator. Next, substitute the given limit. Whenever, there are two polynomials of same degree in numerator and denominator, and limit tends to infinite, take the highest power of the polynomial as common from both numerator and denominator.

Complete step by step solution:
The given limit is shown below.
$\mathop {\lim }\limits_{x \to \infty } \dfrac{{\sqrt 7 {x^2} + 3x - 2}}{{{x^2} + 5}}$
Take ${x^2}$ common from both numerator and denominator and solve the above expression.
$\,\,\,\,\,\,\mathop {\lim }\limits_{x \to \infty } \dfrac{{{x^2}\left( {\sqrt 7 + \dfrac{3}{x} - \dfrac{2}{{{x^2}}}} \right)}}{{{x^2}\left( {1 + \dfrac{5}{{{x^2}}}} \right)}} \\ \Rightarrow \mathop {\lim }\limits_{x \to \infty } \dfrac{{\sqrt 7 + \dfrac{3}{x} - \dfrac{2}{{{x^2}}}}}{{1 + \dfrac{5}{{{x^2}}}}} \\ \Rightarrow \dfrac{{\sqrt 7 + 0 - 0}}{{1 + 0}} \\ \Rightarrow \sqrt 7 \\ \Rightarrow \sqrt 7 \\ \Rightarrow 2.65 \\$

Thus, the value of the expression $\mathop {\lim }\limits_{x \to \infty } \dfrac{{\sqrt 7 {x^2} + 3x - 2}}{{{x^2} + 5}}$ is 2.65, hence, option (D) is correct answer.

Note: Take highest power of x as common from both numerator and denominator. We can solve the given problem, using L. hospital rules also. The L hospital rule says that, when the given expression is in the form of $\dfrac{0}{0}\,\,\,{\text{or}}\,\,\,\dfrac{\infty }{\infty }$, after substituting the given limits, differentiate the numerator and denominator of the given expression.