Calculus: Limits & Continuity

As x gets closer and closer to a (from both sides), f(x) gets arbitrarily close to L . The value of f(a) itself does not matter.

When both one-sided limits exist and are equal: \displaystyle\lim_{x \to a^-} f(x) = \lim_{x \to a^+} f(x)

1. f(a) is defined 2. \displaystyle\lim_{x \to a} f(x) exists 3. \displaystyle\lim_{x \to a} f(x) = f(a)

Removable: a hole (the limit exists but does not equal f(a)) Jump: left and right limits differ Infinite: a vertical asymptote

1

0

1

e \approx 2.71828

5

6

Multiply by the conjugate \sqrt{x} + 2 : \dfrac{x - 4}{(x - 4)(\sqrt{x} + 2)} = \dfrac{1}{\sqrt{x} + 2} \to \dfrac{1}{4}

It is an indeterminate form, not an answer. Try factoring and canceling, multiplying by a conjugate, simplifying a complex fraction, or L'Hopital's rule.

If \dfrac{f(x)}{g(x)} gives \tfrac{0}{0} or \tfrac{\infty}{\infty} , then \displaystyle\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)}

Top degree < bottom degree: limit is 0 Degrees equal: ratio of leading coefficients Top degree > bottom degree: limit is ±∞ (no horizontal asymptote)

3/5

It does not exist: \displaystyle\lim_{x \to 0^+} \frac{1}{x} = +\infty but \displaystyle\lim_{x \to 0^-} \frac{1}{x} = -\infty .

Vertical asymptote x = a : f(x) \to \pm\infty as x \to a . Horizontal asymptote y = L : f(x) \to L as x \to \pm\infty .

If g(x) \le f(x) \le h(x) near a and \displaystyle\lim_{x \to a} g(x) = \lim_{x \to a} h(x) = L , then \displaystyle\lim_{x \to a} f(x) = L .

If f is continuous on [a, b] and N is between f(a) and f(b) , then f(c) = N for at least one c in (a, b) .

For every \varepsilon > 0 there is a \delta > 0 such that 0 < |x - a| < \delta implies |f(x) - L| < \varepsilon .