Question

$\underset{x\to 2}{lim}\frac{x^2+5x+6}{2x^2-3x}$

• A. $10$
• B. $\infty$
• C. $2$
• D. $5$

Answer 1

The correct option is A $10$

Both the denominator and the numerator are non-zero algebraic polynomials,

so we can directly put $x=2$

$\therefore \underset{x⟶2}{lim}\frac{x^2+5x+6}{{2x}^2-3x}$

$=\frac{2^2+5\times 2+6}{2\times 2^2-3\times 2}$

$=\frac{4+10+6}{8-6}$ =$\frac{20}{2}=10$