Evaluating a Simple Limit

Find $\lim_{x \to 3} (2x + 1)$

Check if direct substitution works: The function $f(x) = 2x + 1$ is defined at $x = 3$ = Direct substitution is valid

Substitute the value: $f(3) = 2(3) + 1 = 6 + 1$ = $= 7$

Write the answer in limit notation: $\lim_{x \to 3} (2x + 1) = 7$ = The limit is $7$

Answer: $\lim_{x \to 3} (2x + 1) = 7$

Limit with a Hole (Factoring)

Find $\lim_{x \to 2} \frac{x^2 - 4}{x - 2}$

Try direct substitution: $\frac{2^2 - 4}{2 - 2} = \frac{4 - 4}{0} = \frac{0}{0}$ = Indeterminate form — need another approach

Factor the numerator: $x^2 - 4 = (x+2)(x-2)$ = Difference of squares

Simplify the fraction: $\frac{(x+2)(x-2)}{x-2} = x + 2$ (for $x \neq 2$) = Cancel common factor

Now substitute: $\lim_{x \to 2} (x + 2) = 2 + 2$ = $= 4$

Answer: $\lim_{x \to 2} \frac{x^2 - 4}{x - 2} = 4$

One-Sided Limits

Find $\lim_{x \to 0^+} \frac{1}{x}$ and $\lim_{x \to 0^-} \frac{1}{x}$

Understand one-sided notation: $x \to 0^+$ means approaching from the right (positive side) $x \to 0^-$ means approaching from the left (negative side) = Two separate limits to evaluate

Evaluate from the right: As $x \to 0^+$: $x$ is small and positive, so $\frac{1}{x}$ is large and positive = $\lim_{x \to 0^+} \frac{1}{x} = +\infty$

Evaluate from the left: As $x \to 0^-$: $x$ is small and negative, so $\frac{1}{x}$ is large and negative = $\lim_{x \to 0^-} \frac{1}{x} = -\infty$

Conclude about the two-sided limit: Since left and right limits differ, the two-sided limit does not exist = $\lim_{x \to 0} \frac{1}{x}$ DNE

Answer: $\lim_{x \to 0^+} \frac{1}{x} = +\infty$, $\lim_{x \to 0^-} \frac{1}{x} = -\infty$, two-sided limit does not exist

Limit at Infinity

Find $\lim_{x \to \infty} \frac{3x^2 + 2}{x^2 - 1}$

Identify the dominant terms: Both numerator and denominator have degree 2 = Compare leading coefficients

Divide by highest power: $\frac{3x^2 + 2}{x^2 - 1} = \frac{3 + \frac{2}{x^2}}{1 - \frac{1}{x^2}}$ = Divide all terms by $x^2$

Take the limit: As $x \to \infty$: $\frac{2}{x^2} \to 0$ and $\frac{1}{x^2} \to 0$ = $\frac{3 + 0}{1 - 0} = 3$

State the result: $\lim_{x \to \infty} \frac{3x^2 + 2}{x^2 - 1} = 3$ = Horizontal asymptote at $y = 3$

Answer: $\lim_{x \to \infty} \frac{3x^2 + 2}{x^2 - 1} = 3$

Mistake: Confusing $f(a)$ with $\lim_{x \to a} f(x)$

Why: The value of a function AT a point can differ from (or not exist while) the limit EXISTS. Limits describe approaching behavior, not the actual value.

Correct: Always think: 'What does $f(x)$ get close to?' not 'What is $f(a)$?'

Mistake: Saying a limit 'equals infinity' means it exists

Why: When we write $\lim = \infty$, we're describing unbounded growth. Technically, the limit 'does not exist' as a finite number, but we use infinity notation to describe the behavior.

Correct: Distinguish between: DNE (doesn't exist), $= L$ (exists, equals $L$), $= \infty$ (unbounded)

Mistake: Forgetting to check both sides for existence

Why: A two-sided limit exists only if both one-sided limits exist AND are equal.

Correct: $\lim_{x \to a} f(x) = L$ requires $\lim_{x \to a^-} f(x) = \lim_{x \to a^+} f(x) = L$

Mistake: Canceling $\frac{0}{0}$ incorrectly

Why: $\frac{0}{0}$ is indeterminate, not equal to 1 or 0. It signals that algebraic manipulation is needed.

Correct: When you get $\frac{0}{0}$, factor, rationalize, or use other techniques to simplify first.

Instantaneous Speed

When you check your speedometer, you see your instantaneous speed — not your average speed. This is calculated using limits.

If your position is $s(t) = t^2$ meters at time $t$ seconds, your instantaneous speed at $t = 3$ is $\lim_{h \to 0} \frac{s(3+h) - s(3)}{h} = \lim_{h \to 0} \frac{(3+h)^2 - 9}{h} = 6$ m/s

Population Growth Models

Biologists use limits to model carrying capacity — the maximum population an environment can sustain.

The logistic model $P(t) = \frac{K}{1 + Ae^{-rt}}$ shows that $\lim_{t \to \infty} P(t) = K$, the carrying capacity.

Compound Interest and $e$

The number $e \approx 2.718$ comes from a limit involving compound interest calculated infinitely often.

$e = \lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^n$ represents continuous compounding.

Q: What's the difference between a limit and the function value?

A: The function value $f(a)$ is the output when you plug in $a$. The limit $\lim_{x \to a} f(x)$ describes what the outputs approach as inputs get close to $a$. These can be different! For example, a function might have a hole at $x = 2$ (so $f(2)$ is undefined) but still have a limit there.

Q: What does 'does not exist' mean for a limit?

A: A limit DNE when there's no single value that $f(x)$ approaches. This happens when: (1) left and right limits are different, (2) the function oscillates wildly, or (3) we say '$= \infty$' to indicate unbounded growth (technically DNE as a finite value).

Q: Why is $\frac{0}{0}$ called 'indeterminate'?

A: Because $\frac{0}{0}$ doesn't determine a specific value — the actual limit could be any number or even infinity. It just signals that the numerator and denominator both vanish, and you need to dig deeper (factor, simplify, etc.) to find the true limit.