Question

The value of the function is
$f\left(x\right)={\text{lim}}_{x\to 0}\frac{2x^3+3x^2}{4x^3-5x^2}$

• A. $\frac{-3}{5}$
• B. 0
• C. $\infty$
• D. $\frac{3}{5}$

Answer 1

The correct option is A $\frac{-3}{5}$

$f\left(x\right)={\text{lim}}_{x\to 0}\left[\frac{2x^3+3x^2}{4x^3-5x^2}\right]$
Since this has $\frac{0}{0}$ form, limit can be found by repeated application of L'Hospitals rule.
$f\left(x\right)={\text{lim}}_{x\to 0}\left[\frac{6x^2+6x}{12x^2-10x}\right]$
$={\text{lim}}_{x\to 0}\left[\frac{12x+6}{24x-10}\right]=\left[\frac{12\times 0+6}{24\times 0-10}\right]=\frac{-6}{10}$
$=\frac{-3}{5}$