Solveeit Logo

Question

Question: \(\lim_{x \rightarrow \infty}\frac{ax^{2} + bx + c}{dx^{2} + ex + f}\)=...

limxax2+bx+cdx2+ex+f\lim_{x \rightarrow \infty}\frac{ax^{2} + bx + c}{dx^{2} + ex + f}=

A

be\frac{b}{e}

B

cf\frac{c}{f}

C

ad\frac{a}{d}

D

da\frac{d}{a}

Answer

ad\frac{a}{d}

Explanation

Solution

Here the expression assumes the form \frac{\infty}{\infty}. We note that the highest power of x in both the numerator and denominator is 2. So we divide each terms in both the numerator and denominator by x2x^{2}.

limxax2+bx+cdx2+ex+f=limxa+bx+cx2d+ex+fx2=a+0+0d+0+0=ad\lim_{x \rightarrow \infty}\frac{ax^{2} + bx + c}{dx^{2} + ex + f} = \lim_{x \rightarrow \infty}\frac{a + \frac{b}{x} + \frac{c}{x^{2}}}{d + \frac{e}{x} + \frac{f}{x^{2}}} = \frac{a + 0 + 0}{d + 0 + 0} = \frac{a}{d}.