Limits describe what a function is approaching, not necessarily what it equals.
Calculus is built on the foundation of limits. Think about driving a car: your speedometer shows your instantaneous speed at any moment, but speed is actually distance divided by time. How can we measure speed at a single instant if time isn't passing?
This is where limits come in. Limits help us describe what happens as we get infinitely close to a point, even if we can't evaluate the function exactly at that point. They're the mathematical tool that lets us:
We can't always plug in the exact value because it may "break" the function (like division by zero), but we can study the trend as we get arbitrarily close.
The formal definition of a limit is written as:
This notation means:
"As x gets closer and closer to a, the values of f(x) get closer and closer to L."
Consider the function $f(x) = x^2$. As $x$ approaches $2$, what does $f(x)$ approach?
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Notice how as $x$ gets closer to $2$ (from either side), the function values get closer to $4$. The dashed lines show $x = 2$ and the limit value $L = 4$.
Important distinction:
Sometimes a function behaves differently when we approach from the left versus the right. We use special notation for this:
Approaching from values less than $a$ (from the left on a number line)
Approaching from values greater than $a$ (from the right on a number line)
For $f(x) = x^2$ at $x = 2$, both the left and right limits equal $4$:
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Since $\lim_{x \to 2^-} f(x) = \lim_{x \to 2^+} f(x) = 4$, the two-sided limit exists and equals $4$.
Consider a piecewise function where the left and right limits differ. For example, at $x = 0$:
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Here, $\lim_{x \to 0^-} f(x) = 2$ but $\lim_{x \to 0^+} f(x) = 1$. Since they don't match,$\lim_{x \to 0} f(x)$ does not exist.
Decision Rule for Two-Sided Limits:
✓ If $\lim_{x \to a^-} f(x) = \lim_{x \to a^+} f(x) = L$, then $\lim_{x \to a} f(x) = L$ (limit exists)
✗ If $\lim_{x \to a^-} f(x) \neq \lim_{x \to a^+} f(x)$, then $\lim_{x \to a} f(x)$ does not exist
Let's work through a classic limit problem that illustrates why limits are so useful:
If we plug in $x = -6$ directly:
We get $\frac{0}{0}$ , which is indeterminate. This means we need to try algebra to simplify the expression.
We need two numbers that multiply to $-12$ and add to $4$. Those numbers are $6$ and $-2$.
$$x^2 + 4x - 12 = (x + 6)(x - 2)$$
$$\frac{x^2 + 4x - 12}{x + 6} = \frac{(x + 6)(x - 2)}{x + 6} = x - 2$$ (for $x \neq -6$)
The restriction "$x \neq -6$" is crucial. We can cancel the $(x + 6)$ terms, but we must remember that $x = -6$ creates a "hole" in the graph — the function is undefined there.
Now we can find the limit by substituting $x = -6$ into the simplified form:
Here's what the graph of $f(x) = \frac{x^2 + 4x - 12}{x + 6} = x - 2$ (for $x \neq -6$) looks like:
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Notice the hole at $x = -6$. Even though the function is undefined at that point, the limit exists and equals $-8$ because the function approaches $-8$ from both sides.
| x | f(x) = x - 2 | Approaching from |
|---|---|---|
| -6.1 | -8.1 | Left |
| -6.01 | -8.01 | Left |
| -6.001 | -8.001 | Left |
| -6 | undefined | The hole |
| -5.999 | -7.999 | Right |
| -5.99 | -7.99 | Right |
| -5.9 | -7.9 | Right |
As $x$ approaches $-6$ from both sides, $f(x)$ approaches $-8$, confirming our answer.
Final Answer:
It's important to be able to translate between words and mathematical notation. Let's practice:
Statement in words:
"As $$x$$ gets close to $$1$$, the function $$y = x + 2$$ approaches $$3$$"
In limit notation:
1. "As $$x$$ approaches $$5$$, the function $$f(x) = 2x$$ approaches $$10$$"
2. "As $$x$$ gets closer to $$0$$, $$g(x) = x^2 + 1$$ gets closer to $$1$$"
Here's a systematic approach to finding limits at this level:
Plug in the value of $x$ directly into the function.
If it works (you get a real number):
✓ That's your answer! You're done.
If it doesn't work (you get $\frac{0}{0}$ or undefined):
→ Move to Step 2
a) Factor and Cancel (most common)
When you get $\frac{0}{0}$, try factoring the numerator and denominator:
b) Rationalize (for square roots)
Multiply by the conjugate to eliminate the square root:
If algebra doesn't resolve it, you'll need more advanced methods like L'Hôpital's Rule or squeeze theorem, which you'll learn later in calculus.
Pro tip:
Always try direct substitution first! About 70% of intro calculus limit problems can be solved this way. Only use algebra when you get an indeterminate form like $\frac{0}{0}$.
Assuming the limit equals the function value
$\lim_{x \to a} f(x)$ can exist even when $f(a)$ is undefined or different
Canceling without restrictions
When you cancel $(x-a)$, remember to note "for $x \neq a$" — this is where the hole exists
Not checking both one-sided limits
For piecewise functions or functions with jumps, always verify left and right limits match
Saying $\frac{0}{0}$ equals 0 or is simply undefined
$\frac{0}{0}$ is indeterminate — it means you need to simplify algebraically, not that the answer is 0 or that the limit doesn't exist
Confusing $\frac{0}{0}$ with $\frac{n}{0}$ where $n \neq 0$
$\frac{5}{0}$ means the limit does not exist (or is infinite), while $\frac{0}{0}$ means you need to simplify
Work through these problems to master the concepts. Reveal hints if you get stuck, and check the full solutions to verify your work.
Test your understanding with these multiple-choice questions. Select your answer and check to see if you're correct.
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