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Calculus

Unit 1

Limits

6 topics · 24 practice questions · Free responseNot yet reviewed

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Notes

The idea of a limit

  • $\displaystyle \lim_{x \to a} f(x) = L$ means $f(x)$ gets close to $L$ as $x$ gets close to $a$. It depends on values near $a$, not on $f(a)$.
  • Estimate limits from tables (inputs closer and closer to $a$ from both sides) or graphs (follow the curve toward $a$).
  • A two-sided limit exists only when the left-hand and right-hand limits exist and are equal.

Computing limits

  • Limit laws let you add, subtract, multiply and divide limits (divide only if the bottom limit is not 0).
  • Try direct substitution first. If you get $\displaystyle \frac{0}{0}$, factor and cancel or multiply by a conjugate.
  • A nonzero number over 0 means the function grows without bound, so the limit does not exist as a finite number.

Limits at infinity

  • For rational functions compare degrees: smaller top gives 0, equal degrees give the ratio of leading coefficients, larger top grows without bound.
  • A finite limit at infinity gives a horizontal asymptote.

Topics in this unit

Each topic has its own short notes and practice.

Practice

Find $\displaystyle \lim_{x \to 4} \frac{x^2 - 16}{x - 4}$.
As $x \to 2^-$, $f(x) \to 1$. As $x \to 2^+$, $f(x) \to 4$. Which statement is correct?
Find $\displaystyle \lim_{x \to \infty} \frac{5x^2 - 1}{2x^2 + x}$.
Find $\displaystyle \lim_{x \to 0} \frac{\sqrt{x + 9} - 3}{x}$.
If $\displaystyle \lim_{x \to 1} f(x) = 4$ and $\displaystyle \lim_{x \to 1} g(x) = -1$, find $\displaystyle \lim_{x \to 1} \left(f(x) g(x) + 3\right)$.
Which statement about limits is false?

Free response

Write your answer first, then open the scoring guide.

Free-response question

Let $f(x) = \frac{x^2 - x - 6}{x - 3}$ and $g(x) = \frac{5x^2 + 2x}{2x^2 - 1}$.

(a) Find $\lim_{x \to 3} f(x)$.

Show the answer and scoring guide

Factor the top: $x^2 - x - 6 = (x - 3)(x + 2)$. For $x \neq 3$, $f(x) = x + 2$, so the limit is 5.

Check off the points you earned:

(b) Is $f(3)$ defined? Explain how this relates to your answer in (a).

Show the answer and scoring guide

No, the denominator is 0 at $x = 3$, so $f(3)$ is undefined. The limit still exists because it depends only on values near 3.

Check off the points you earned:

(c) Find $\lim_{x \to \infty} g(x)$ and state the horizontal asymptote.

Show the answer and scoring guide

The degrees are equal, so the limit is the ratio of leading coefficients: $\displaystyle \frac{5}{2}$. The horizontal asymptote is $\displaystyle y = \frac{5}{2}$.

Check off the points you earned: