Answer
$\dfrac{1}{4}$
Work Step by Step
Our aim is to compute the value of $\lim\limits_{x \to \infty} \dfrac{x^2-x-6}{4x^2-3}$.
Here, we need to use the formula of:
$\lim \limits_{x\to a}f(x)=f(a)$, this implies that:
$\lim\limits_{x \to \infty} \dfrac{x^2-x-6}{4x^2-3}=\dfrac{1-\dfrac{1}{x}-\dfrac{6}{x^2}}{4-\dfrac{3}{x^2}}\\=\dfrac{1-\dfrac{1}{\infty}-\dfrac{6}{(\infty)^2}}{4-\dfrac{3}{(\infty)^2}}\\=\dfrac{1-0-0}{4-0}\\=\dfrac{1}{4}$
