Foundations Reference

Limits

A limit describes the value a function approaches as its input approaches a target. The function may be undefined at the target and still have a limit there.

Solved Examples

Jump directly to a worked example by problem type.

Limit Theorems

Let $\lim_{x\to a}f(x)=L$ and $\lim_{x\to a}g(x)=M$. The following rules let you build new limits from known limits.

TheoremStatementCondition or Use
1. Constant$\lim_{x\to a}c=c$A fixed value does not change.
2. Identity$\lim_{x\to a}x=a$The input approaches the target.
3. Sum$\lim(f+g)=L+M$Add the separate limits.
4. Difference$\lim(f-g)=L-M$Subtract the separate limits.
5. Constant Multiple$\lim(cf)=cL$Pull a constant outside the limit.
6. Product$\lim(fg)=LM$Multiply the separate limits.
7. Quotient$\lim\frac{f}{g}=\frac{L}{M}$Requires $M\ne0$.
8. Positive Integer Power$\lim[f(x)]^n=L^n$For $n=1,2,3,\ldots$.
9. Root$\lim\sqrt[n]{f(x)}=\sqrt[n]{L}$Use when the root is defined near $a$.
10. Polynomial$\lim P(x)=P(a)$Every polynomial is continuous everywhere.
11. Rational Function$\lim\frac{P(x)}{Q(x)}=\frac{P(a)}{Q(a)}$Requires $Q(a)\ne0$.
12. Composition$\lim f(g(x))=f(M)$Use when $f$ is continuous at $M$.
The product, quotient, power, and root theorems are the main tools for evaluating expressions built from simpler limits. If a denominator becomes zero, or a root is not defined near the target, check the conditions before applying the rule.

Test Direct Substitution

For polynomials, rational functions with a nonzero denominator, and other functions continuous at $a$, substitute $x=a$.

Polynomial
$$\lim_{x\to2}(x^2+3x-1)=2^2+3(2)-1=9$$
Rational Function
$$\lim_{x\to1}\frac{x+2}{x+3}=\frac{1+2}{1+3}=\frac34$$
If direct substitution gives a real number, the limit equals that value. Stop there. A complicated-looking limit does not automatically require a special technique.

One Complete Limit-Law Example

Evaluate every part of the expression before combining the results:

Calculate All Component Limits

Find $$\lim_{x\to4}\frac{x^2+3x-1}{\sqrt{x+5}}.$$

Solution Steps
  1. Test direct substitution by setting $x=4$ in the original expression.
  2. Evaluate the power terms at $x=4$.
  3. Evaluate the numerator using the sum and difference laws.
  4. Evaluate the expression under the radical at $x=4$.
  5. Evaluate the square root.
  6. Apply the quotient law because the denominator limit is nonzero.
  7. State the original limit: $\lim_{x\to4}\frac{x^2+3x-1}{\sqrt{x+5}}=9$.
$$\lim_{x\to4}x^2=4^2=16$$$$\lim_{x\to4}3x=3(4)=12$$$$\lim_{x\to4}1=1$$$$\lim_{x\to4}(x^2+3x-1)=16+12-1=27$$$$\lim_{x\to4}(x+5)=4+5=9$$$$\lim_{x\to4}\sqrt{x+5}=\sqrt9=3$$$$\lim_{x\to4}\frac{x^2+3x-1}{\sqrt{x+5}}=\frac{27}{3}=9$$
The final quotient law is valid because the denominator limit is $3$, not $0$. This is the full limit-law process: calculate the pieces, combine the numerator, calculate the denominator, then divide.

When Substitution Gives $0/0$

The form $\frac00$ is indeterminate: it tells you to simplify before evaluating, not that the limit is zero.

Factor and Cancel

Find $\lim_{x\to3}\frac{x^2-9}{x-3}$.

Solution Steps
  1. Test direct substitution at $x=3$; the original expression evaluates to $0/0$.
  2. Factor the numerator and cancel the common factor for $x\ne3$.
  3. Evaluate the simplified limit and state the original limit: $\lim_{x\to3}\frac{x^2-9}{x-3}=6$.
$$\frac{x^2-9}{x-3}=\frac{(x-3)(x+3)}{x-3}$$$$=x+3\quad (x\ne3)$$$$\lim_{x\to3}\frac{x^2-9}{x-3}=6$$

Rationalize

Find $\lim_{x\to0}\frac{\sqrt{x+1}-1}{x}$.

Solution Steps
  1. Test direct substitution at $x=0$; the original expression evaluates to $0/0$.
  2. Multiply by the conjugate, use difference of squares, and cancel $x$.
  3. Evaluate the simplified limit and state the original limit: $\lim_{x\to0}\frac{\sqrt{x+1}-1}{x}=\frac12$.
$$\frac{\sqrt{x+1}-1}{x}\cdot\frac{\sqrt{x+1}+1}{\sqrt{x+1}+1}$$$$=\frac{1}{\sqrt{x+1}+1}$$$$\lim_{x\to0}\frac{\sqrt{x+1}-1}{x}=\frac12$$

One-Sided Limits

NotationMeaningTwo-Sided Limit
$\lim_{x\to a^-}f(x)$Approach $a$ from values less than $a$.Exists only when the left and right limits agree.
$\lim_{x\to a^+}f(x)$Approach $a$ from values greater than $a$.
If the one-sided limits are different, then $\lim_{x\to a}f(x)$ does not exist, even if both one-sided limits exist.

Limit with Absolute Value

Absolute value can hide a piecewise rule. Check the expression on each side of the target.

Use One-Sided Behavior

Find $\lim_{x\to0}\frac{|x|}{x}.$

Solution Steps
  1. For $x<0$, rewrite $|x|$ as $-x$ and evaluate the left-hand limit.
  2. For $x>0$, rewrite $|x|$ as $x$ and evaluate the right-hand limit.
  3. Compare the one-sided limits: $-1\ne1$.
  4. State the final result: $\lim_{x\to0}\frac{|x|}{x}$ does not exist.
$x<0:\quad \frac{|x|}{x}=\frac{-x}{x}=-1$$x>0:\quad \frac{|x|}{x}=\frac{x}{x}=1$$\lim_{x\to0^-}\frac{|x|}{x}=-1,\qquad \lim_{x\to0^+}\frac{|x|}{x}=1$$\therefore\ \lim_{x\to0}\frac{|x|}{x}\text{ does not exist}$

Continuity

A function is continuous at $x=a$ when all three conditions hold:

$$f(a)\text{ exists},\qquad \lim_{x\to a}f(x)\text{ exists},\qquad \lim_{x\to a}f(x)=f(a)$$

Polynomials are continuous everywhere. Rational functions are continuous wherever their denominators are nonzero. A removable hole can often be repaired by defining the missing point to equal the limit.

Behavior at Infinity

SituationWhat to inspectTypical conclusion
$x\to\infty$ for a rational functionCompare numerator and denominator degrees.Equal degrees approach the ratio of leading coefficients.
Vertical asymptoteApproach a value excluded from the domain.Function values may grow without bound.
Unbounded behaviorUse $\infty$ as a direction, not a real number.$\lim f(x)=\infty$ means values eventually exceed every bound.

Quick Checklist

  1. Identify the target $a$ and the direction of approach.
  2. Try direct substitution.
  3. If the result is $0/0$, factor, cancel, combine, or rationalize.
  4. Check one-sided behavior when the function has a break, denominator zero, or piecewise definition.
  5. State whether the limit is finite, infinite, or does not exist.