Computing Limits Algebraically

When evaluating limits in calculus, looking at a graph isn't always practical or precise. Computing limits algebraically allows you to find the exact value of a limit using algebraic manipulation and established limit laws.

Direct Substitution

The first step in evaluating any limit is to simply plug the target value into the function. If the function is continuous at that point, direct substitution will give you the answer.

For example, to find $\lim_{x \to 2} (3x^2 - 4x + 1)$: $\lim_{x \to 2} (3x^2 - 4x + 1) = 3(2)^2 - 4(2) + 1 = 12 - 8 + 1 = 5$

Factoring and Canceling

Often, direct substitution results in the indeterminate form $\frac{0}{0}$. When this happens with polynomials, you can usually factor the numerator and denominator to cancel out the problematic term.

Example: Find $\lim_{x \to 3} \frac{x^2 - 9}{x - 3}$

Plugging in $x = 3$ gives $\frac{0}{0}$. Instead, factor the difference of squares in the numerator: $\lim_{x \to 3} \frac{(x - 3)(x + 3)}{x - 3}$ Cancel the common $(x - 3)$ term: $\lim_{x \to 3} (x + 3)$ Now, use direct substitution: $3 + 3 = 6$

Rationalizing

If a limit yields $\frac{0}{0}$ and contains a square root, multiplying the numerator and denominator by the conjugate of the radical expression will often clear the issue.

For example, if you have $\frac{\sqrt{x} - 2}{x - 4}$ as $x \to 4$, you multiply the top and bottom by the conjugate, $(\sqrt{x} + 2)$, to eliminate the root in the numerator and cancel the $(x - 4)$ term.

Special Trigonometric Limits

Some limits cannot be evaluated using basic algebra and rely on special established limits. The most important one to memorize is: $\lim_{x \to 0} \frac{\sin x}{x} = 1$

You can use algebraic manipulation to make a given problem match this special form.

Example: Find $\lim_{x \to 0} \frac{\sin(5x)}{3x}$

To use the special limit, the angle inside the sine function must exactly match the denominator. We need a $5x$ in the denominator.

Multiply the numerator and denominator by $5$: $\lim_{x \to 0} \frac{5 \cdot \sin(5x)}{5 \cdot 3x}$ Rearrange the constants: $\lim_{x \to 0} \frac{5}{3} \cdot \frac{\sin(5x)}{5x}$ Since $\lim_{x \to 0} \frac{\sin(5x)}{5x} = 1$, we get: $\frac{5}{3} \cdot 1 = \frac{5}{3}$

Limit Laws

When computing limits algebraically, you are relying on fundamental limit laws. Assuming $\lim_{x \to a} f(x)$ and $\lim_{x \to a} g(x)$ both exist, you can use:

• Sum/Difference Law: $\lim_{x \to a} [f(x) \pm g(x)] = \lim_{x \to a} f(x) \pm \lim_{x \to a} g(x)$
• Product Law: $\lim_{x \to a} [f(x) \cdot g(x)] = \lim_{x \to a} f(x) \cdot \lim_{x \to a} g(x)$
• Quotient Law: $\lim_{x \to a} \frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}$ (provided the denominator's limit is not zero)