Every way to evaluate a limit
What to try first, and what to reach for when the obvious approach fails.
Try direct substitution first — if the function is continuous at the point, that's the whole problem. If substitution gives 0/0 (or \infty/\infty), the function is a quotient, and the pieces are differentiable, L'Hopital's Rule is usually the fastest fix. If it isn't a quotient, or L'Hopital's doesn't apply, go back to algebra: factor, rationalize, or simplify until the 0/0 cancels on its own. Reach for the Squeeze Theorem only when the function is trapped between two others you already know the limit of — it isn't a general-purpose tool. A one-sided check settles anything with a piecewise definition or an absolute value; a limit at infinity needs its own end-behavior reasoning, not any of the point-limit techniques.
Start here
Term
continuity
If f is continuous at a, \lim_{x\to a} f(x) = f(a) — substitution IS the answer.
Theorem
Limit laws
Suppose \lim_{x \to a} f(x) = L and \lim_{x \to a} g(x) = M, both real numbers, and c is a constant. Then:
- Sum/difference: \displaystyle\lim_{x \to a} [f(x) \pm g(x)] = L \pm M
- Constant multiple: \displaystyle\lim_{x \to a} [c \cdot f(x)] = cL
- Product: \displaystyle\lim_{x \to a} [f(x) \cdot g(x)] = LM
- Quotient: \displaystyle\lim_{x \to a} \frac{f(x)}{g(x)} = \frac{L}{M}, provided M \neq 0
- Power: \displaystyle\lim_{x \to a} [f(x)]^n = L^n for any positive integer n
- Root: \displaystyle\lim_{x \to a} \sqrt[n]{f(x)} = \sqrt[n]{L}, provided L \geq 0 when n is even
Combine known limits of pieces (sum, product, quotient) instead of re-deriving from scratch.
Show proof
Each law reduces to the same style of argument used to make a limit claim rigorous (see this course's section on the precise definition of a limit): to bound |f(x)+g(x) - (L+M)|, use the triangle inequality to split it as |f(x)-L| + |g(x)-M|, then make each piece small enough by staying close enough to a to satisfy both limit definitions at once. The product and quotient laws need one more step — bounding f(x) near a so a factor of g(x)-M doesn't get multiplied by something unbounded — but the shape of the argument is identical. This is the precise version of the intuition "if two things are close to their targets, arithmetic combinations of them are close to the combined target."
Substitution gives 0/0 or an undefined form
Suppose f and g are differentiable near a (except possibly at a), g'(x) \neq 0 near a, and \lim_{x\to a}\frac{f(x)}{g(x)} \text{ is of the indeterminate form } \frac{0}{0} \text{ or } \frac{\pm\infty}{\pm\infty}. Then \lim_{x\to a}\frac{f(x)}{g(x)} = \lim_{x\to a}\frac{f'(x)}{g'(x)}, provided the limit on the right exists (or is \pm\infty). The rule also holds for x \to a^+, x \to a^-, or x \to \pm\infty.
Example
For a quotient stuck at 0/0 or \pm\infty/\pm\infty — replace top and bottom with their derivatives.
Show proof
The \frac{0}{0} case follows from the Generalized (Cauchy) Mean Value Theorem: for x near a, there is a c between a and x with \dfrac{f(x)-f(a)}{g(x)-g(a)} = \dfrac{f'(c)}{g'(c)}. Since f(a)=g(a)=0 in the \frac{0}{0} case, the left side is exactly \frac{f(x)}{g(x)}; letting x \to a forces c \to a too, giving the limit of f'/g'. The \frac{\infty}{\infty} case uses the same idea with a more careful algebraic argument and is not reproduced here.
The function is trapped between two others
Example
Piecewise definitions, or the input is unbounded
Check the left- and right-hand limits separately at a break point.
Divide by the highest power present, or compare growth rates directly.