Calculus - Limits (HL)

Evaluate the limit: $\lim_{x \to 3} \frac{x^2 - 9}{x - 3}$.

$\lim_{x \to 3} \frac{(x - 3)(x + 3)}{x - 3} = \lim_{x \to 3} (x + 3) = 3 + 3 = 6$

Direct substitution results in the indeterminate form $\frac{0}{0}$. We factor the numerator using the difference of squares, cancel the common factor $(x - 3)$, and then substitute $x = 3$.

Find the horizontal asymptote of the function $f(x) = \frac{4x^2 + 5}{2x^2 - 3x}$ by evaluating the limit at infinity.

$\lim_{x \to \infty} \frac{4x^2 + 5}{2x^2 - 3x} = \lim_{x \to \infty} \frac{\frac{4x^2}{x^2} + \frac{5}{x^2}}{\frac{2x^2}{x^2} - \frac{3x}{x^2}} = \lim_{x \to \infty} \frac{4 + \frac{5}{x^2}}{2 - \frac{3}{x}} = \frac{4 + 0}{2 - 0} = 2$

To find the limit at infinity for a rational function, divide every term by the highest power of $x$ in the denominator ($x^2$). As $x \to \infty$, terms like $\frac{5}{x^2}$ and $\frac{3}{x}$ approach zero.

Use L'Hôpital's Rule to evaluate $\lim_{x \to 0} \frac{e^{2x} - 1}{x}$.

Let $f(x) = e^{2x} - 1$ and $g(x) = x$. Since $f(0) = 0$ and $g(0) = 0$, we apply L'Hôpital's Rule: $\lim_{x \to 0} \frac{f'(x)}{g'(x)} = \lim_{x \to 0} \frac{2e^{2x}}{1} = 2e^0 = 2$

Because the limit results in the indeterminate form $\frac{0}{0}$, we differentiate the numerator ($2e^{2x}$) and the denominator ($1$) and then substitute $x=0$.