Grade 12 · MCV4U · Calculus & Vectors

MCV4U: Evaluating Limits

A limit asks what value $f(x)$ is heading toward as $x$ gets close to a number, not what happens at the number itself. The idea takes a minute; the method is mechanical.

How It Works

Step 1: plug the number in. You will get one of three results, and each tells you what to do next.

A real number. Done, that is the limit. $\lim_{x \to 3} x^2 = 9$ .

A nonzero number over zero, like $\frac{4}{0}$ . The function blows up, so the limit does not exist. Check each side: one may go to $\infty$ and the other to $-\infty$ .

** $\frac{0}{0}$ .** This is indeterminate. It means "not yet", not "zero" and not "does not exist". Something cancels, and your job is to find it.

Step 2: rewrite, then plug in again. Three moves cover almost every $\frac{0}{0}$ in this course:

Factor and cancel. If plugging in $a$ gives $\frac{0}{0}$ , then $(x - a)$ is a factor of the top and the bottom. Factor both and cancel it.

Rationalize. If there is a square root, multiply top and bottom by the conjugate: $\sqrt{x + 4} - 3$ pairs with $\sqrt{x + 4} + 3$ . The difference of squares clears the root and exposes the factor.

Clear the inner fractions. For a fraction inside a fraction, combine the small fractions over a common denominator first.

Cancelling is allowed because a limit only looks near $a$ , never at it, so $x \ne a$ the whole time. That is also why the simplified function can have a value where the original had a hole.

Write $\lim_{x \to a}$ on every line until the last substitution. Dropping it mid-solution says the expressions are equal, and they are not: one is a limit and the other is a function.

5 Worked Examples

Easy first, test level last. Try each one on paper, then open the solution.

Easy
Evaluate $\lim_{x \to 2}(x^2 + 3x - 1)$ .
Show solution
1. Polynomials are continuous, so substitute: $2^2 + 3(2) - 1 = 4 + 6 - 1 = 9$ .
Answer: $9$
Easy
Evaluate $\lim_{x \to 3}\frac{x^2 - 9}{x - 3}$ .
Show solution
1. Substituting gives $\frac{0}{0}$ , so rewrite.
2. Factor the difference of squares: $\lim_{x \to 3}\frac{(x - 3)(x + 3)}{x - 3} = \lim_{x \to 3}(x + 3)$ .
3. Substitute: $3 + 3 = 6$ .
Answer: $6$
Medium
Evaluate $\lim_{x \to -1}\frac{x^2 + 4x + 3}{x^2 - x - 2}$ .
Show solution
1. Substituting gives $\frac{1 - 4 + 3}{1 + 1 - 2} = \frac{0}{0}$ , so $(x + 1)$ is a factor of both.
2. Factor: $\lim_{x \to -1}\frac{(x + 1)(x + 3)}{(x + 1)(x - 2)} = \lim_{x \to -1}\frac{x + 3}{x - 2}$ .
3. Substitute: $\frac{-1 + 3}{-1 - 2} = -\frac{2}{3}$ .
Answer: $-\frac{2}{3}$
Medium
Evaluate $\lim_{x \to 5}\frac{\sqrt{x + 4} - 3}{x - 5}$ .
Show solution
1. Substituting gives $\frac{3 - 3}{0} = \frac{0}{0}$ . There is a root, so rationalize.
2. Multiply top and bottom by $\sqrt{x + 4} + 3$ . The top becomes $(x + 4) - 9 = x - 5$ .
3. $\lim_{x \to 5}\frac{x - 5}{(x - 5)(\sqrt{x + 4} + 3)} = \lim_{x \to 5}\frac{1}{\sqrt{x + 4} + 3}$ .
4. Substitute: $\frac{1}{3 + 3} = \frac{1}{6}$ .
Answer: $\frac{1}{6}$
Test level
Evaluate $\lim_{x \to 3}\frac{\frac{1}{x + 1} - \frac{1}{4}}{x - 3}$ .
Show solution
1. Substituting gives $\frac{\frac{1}{4} - \frac{1}{4}}{0} = \frac{0}{0}$ . Clear the inner fractions first.
2. Common denominator on top: $\frac{1}{x + 1} - \frac{1}{4} = \frac{4 - (x + 1)}{4(x + 1)} = \frac{3 - x}{4(x + 1)}$ .
3. So the expression is $\frac{3 - x}{4(x + 1)(x - 3)}$ . Since $3 - x = -(x - 3)$ , it simplifies to $\frac{-1}{4(x + 1)}$ .
4. Substitute: $\frac{-1}{4(4)} = -\frac{1}{16}$ .
Answer: $-\frac{1}{16}$

Learn It on Martian Lab

Martian Lab teaches this in the Limits lesson of its MCV4U Limits unit. Watch it, then take the Limits practice tests to see if it stuck.

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More MCV4U topics: Chain Rule · Optimization Problems · Related Rates · Curve Sketching Steps · Dot Product and Projections · Cross Product, Area and Torque · Equations of Planes · Intersection of Three Planes

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