CLEP Calculus study materials
CLEP Calculus study resources

CLEP Calculus: Limits and Continuity

CLEP Calculus · Work through your exam, one topic at a time.

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Limits and Continuity

Core ideas

A limit describes the value a function approaches, not necessarily the value it attains. Candidates must be fluent with one-sided limits, two-sided limits, infinite limits (vertical asymptotes), and limits at infinity (horizontal asymptotes and end behavior). Know the limit laws, the Squeeze Theorem, and the special limits $\lim_{x\to0}\frac{\sin x}{x}=1$ and $\lim_{x\to0}\frac{1-\cos x}{x}=0$.

Continuity at a point requires three things: the function is defined there, the limit exists, and the two agree. Distinguish removable discontinuities (a hole), jump discontinuities, and infinite discontinuities. The Intermediate Value Theorem is a favorite: if $f$ is continuous on $[a,b]$ and $k$ lies between $f(a)$ and $f(b)$, some $c$ in $(a,b)$ has $f(c)=k$.

Traps

  • Concluding a limit fails to exist when it is merely a hole; a limit can exist even where the function is undefined.
  • Forgetting that $\lim_{x\to a}$ requires matching one-sided limits.
  • Applying L'Hopital's rule to a limit that is not an indeterminate form like $0/0$ or $\infty/\infty$.
  • Confusing $\lim_{x\to\infty}$ behavior with a vertical asymptote.

Worked example

Evaluate $\lim_{x\to 3}\frac{x^2-9}{x-3}$. Direct substitution gives $0/0$, an indeterminate form. Factor: $\frac{(x-3)(x+3)}{x-3}=x+3$ for $x\neq3$. Thus the limit is $6$. Rule of thumb: try substitution first; if you get $0/0$, factor, rationalize, or simplify before reaching for L'Hopital.