Q.Evaluate $\lim_{x \to 3} (2x^2 - 5x + 1)$.

Substitute $x=3$ directly: the limit equals $3$.

Since the denominator $x+3\neq 0$ at $x=3$, the expression is continuous there and we substitute directly:

$\lim_{x \to 3} \frac{x^2+9}{x+3}=\frac{3^2+9}{3+3}=\frac{9+9}{6}=\frac{18}{6}=3.$

Direct substitution gives $\lim\limits_{x\to 3}(x^2+3)=12$.

$x^2+3$ is a polynomial, hence continuous at every point, so the limit equals the function value at $x=3$:

$\lim_{x\to 3}(x^2+3)=(3)^2+3=9+3=12$

The limit is $2$.

Since the denominator $x+2$ is non-zero at $x = 2$, the function is continuous there and the limit is found by direct substitution:

$\lim_{x \to 2} \frac{x^2 + 4}{x + 2} = \frac{(2)^2 + 4}{2 + 2} = \frac{4 + 4}{4} = \frac{8}{4} = 2.$

Correct option: (a) 4.

At $x = 2$ the expression is $\dfrac{0}{0}$ (indeterminate), so factorise the numerator as a difference of squares:

$\frac{x^2 - 4}{x - 2} = \frac{(x-2)(x+2)}{x - 2} = x + 2 \quad (x \neq 2).$

Now take the limit:

$\lim_{x \to 2} (x + 2) = 2 + 2 = 4.$