Question

$\underset{x\to \infty }{\lim }\left[\frac{x^4+7x^3+46x+a}{x^4+6}\right]$, where a is a non-zero real number.

Answer 1

$\underset{x\to \infty }{\lim }\left[\frac{x^4+7x^3+46x+a}{x^4+6}\right]$

Dividing the numerator and the denominator by x⁴:

$$\begin{gathered} \underset{x\to \infty }{\lim }\left[\frac{1+{\displaystyle \frac{7}{x}}+{\displaystyle \frac{46}{x^2}}+{\displaystyle \frac{9}{x^4}}}{1+{\displaystyle \frac{6}{x^4}}}\right] \\ \mathrm{As}x\to \infty ,\frac{1}{x},\frac{1}{x^2},\frac{1}{x^3},\frac{1}{x^4}\to 0 \\ =\frac{1}{1} \end{gathered}$$