Question

The value of the function f(x) = ${lim}_{x\to 0}\frac{x^3+x^2}{2x^3-7x^2}$ is

• A. Zero
• B. -$\frac{1}{7}$
• C. $\frac{1}{7}$
• D. infinite

Answer 1

The correct option is B -$\frac{1}{7}$

f(x) = ${lim}_{x\to 0}\frac{x^3+x^2}{2x^3-7x^2}$

= ${lim}_{x\to 0}\frac{x^2(x+1)}{x^2(2x-7)}$

= ${lim}_{x\to 0}\left(\frac{x+1}{2x-7}\right)$

= $\frac{0+1}{0-7}=-\frac{1}{7}$