Question

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

Answer 1

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

Dividing numerator and denominator by $x^4,$ we get

$={lim}_{x\to \infty }\frac{1+\frac{7}{x^3}+\frac{46}{x^3}+\frac{a}{x^4}}{1+\frac{6}{x^4}}$

=1