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Limit calculator

One-sided, at infinity, or the indeterminate ones that need real work.

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  • Arithmetic to calculus

A limit asks where a function is heading, not where it lands. Substitution answers most of them immediately. The interesting ones give 0/0 or ∞/∞ on substitution, and that is a signal to do algebra first, not a dead end.

The order of attack

Always substitute first. What comes back tells you which technique is needed.

  1. Substitute

    If you get a number, that is the limit. If you get a non-zero number over zero, the limit is infinite or does not exist, and the one-sided limits decide which.

  2. 0/0 with polynomials: factor and cancel

    The common factor is what caused the zero. Cancel it and substitute again.

  3. 0/0 with a square root: multiply by the conjugate

    Multiplying top and bottom by the conjugate turns the root into a difference of squares and frees the cancellation.

  4. At infinity: divide by the highest power

    Divide top and bottom by the largest power of x in the denominator. Everything with x underneath goes to zero, and the ratio of leading coefficients is left.

  5. L’Hôpital’s rule, only for 0/0 and ∞/∞

    Differentiate the top and bottom separately, then take the limit again. Repeat while the form stays indeterminate.

  6. Check both sides when in doubt

    A limit exists only if the left and right limits agree. At a jump or an absolute value they often do not.

Worked examples

Each one is solved the way the solver solves it: the rule first, then the line.

Find the limit as x approaches 3 of (x^2 - 9)/(x - 3)

lim⁡x→3x2−9x−3\lim_{x \to 3} \frac{x^2 - 9}{x - 3}
  1. Substitution gives an indeterminate form
    9−93−3=00\frac{9 - 9}{3 - 3} = \frac{0}{0}
  2. Factor the numerator
    (x−3)(x+3)x−3\frac{(x - 3)(x + 3)}{x - 3}
  3. Cancel the common factor
    x+3(x≠3)x + 3 \quad (x \ne 3)
  4. Substitute again
    3+3=63 + 3 = 6
Answer
66

Find the limit as x approaches 0 of sin(x)/x

lim⁡x→0sin⁡xx\lim_{x \to 0} \frac{\sin x}{x}
  1. Substitution gives 0/0
    sin⁡00=00\frac{\sin 0}{0} = \frac{0}{0}
  2. Standard limit
    lim⁡x→0sin⁡xx=1\lim_{x \to 0} \frac{\sin x}{x} = 1
  3. Confirm with L’Hôpital
    lim⁡x→0cos⁡x1=1\lim_{x \to 0} \frac{\cos x}{1} = 1
Answer
11

Find the limit as x approaches infinity of (3x^2 + 2x)/(5x^2 - 1)

lim⁡x→∞3x2+2x5x2−1\lim_{x \to \infty} \frac{3x^2 + 2x}{5x^2 - 1}
  1. Divide top and bottom by x²
    3+2x5−1x2\frac{3 + \frac{2}{x}}{5 - \frac{1}{x^2}}
  2. Every term with x underneath tends to zero
    3+05−0\frac{3 + 0}{5 - 0}
  3. Read off the ratio of leading coefficients
    35\frac{3}{5}
Answer
35\dfrac{3}{5}

Find the limit as x approaches 0 of (e^x - 1 - x)/x^2

lim⁡x→0ex−1−xx2\lim_{x \to 0} \frac{e^x - 1 - x}{x^2}
  1. Substitution gives 0/0
    1−1−00=00\frac{1 - 1 - 0}{0} = \frac{0}{0}
  2. L’Hôpital once
    lim⁡x→0ex−12x\lim_{x \to 0} \frac{e^x - 1}{2x}
  3. Still 0/0, so apply it again
    lim⁡x→0ex2\lim_{x \to 0} \frac{e^x}{2}
  4. Substitute
    12\frac{1}{2}
Answer
12\dfrac{1}{2}

Solved problems

Asked here and worked through in full. Open one for every step.

  1. Find the limit as x approaches 3 of (x^2 - 9)/(x - 3)
    Answer 66

Where marks get lost

Using L’Hôpital on the wrong form

It applies only to 0/0 and ∞/∞. On 2/0 or 0·∞ as they stand it gives a wrong answer with full confidence. Rewrite into a quotient first.

Differentiating with the quotient rule

L’Hôpital differentiates the top and the bottom separately. It is not the derivative of the fraction.

Assuming the limit is the value

A function can have a limit at a point where it is undefined, and a value different from its limit. That gap is exactly what continuity is about.

Questions

What is a limit in calculus?

The value a function approaches as the input approaches some point, whether or not the function is defined there. It is what makes derivatives and integrals possible.

How do I find a limit?

Substitute the value first. If you get a number, that is the limit. If you get 0/0, factor, rationalise or simplify until the trouble cancels, then substitute again.

How do I find a limit at infinity?

Compare the highest powers. Divide the top and bottom by the largest power of x present; every term with x underneath goes to zero, leaving the ratio of the leading coefficients.

What is L Hopital rule?

When a limit gives 0/0 or ∞/∞, differentiate the top and the bottom separately and take the limit again. It applies only to those indeterminate forms.

What does a one-sided limit mean?

The value approached from only the left or only the right. If the two sides disagree, the two-sided limit does not exist, which is what happens at a jump.

Why is the limit of sin x over x equal to 1?

Substituting gives 0/0, but the squeeze theorem traps the ratio between two expressions that both approach 1 as x approaches 0. It is the limit that makes the derivative of sine work.

What is an indeterminate form?

An expression like 0/0, ∞/∞ or 0 × ∞ that does not settle the answer on its own. It means more work is needed, not that the limit fails to exist.

Can a limit exist where the function does not?

Yes, and that is the usual case. The function can have a hole at the point while the values on either side still close in on the same number.