SolveMathAI

Limit calculator

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

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  • Every step shown
  • 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}

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

Can it do one-sided limits?

Yes. Say "as x approaches 2 from the left" or write x → 2⁻, and it evaluates that side only.

What does it do when the limit does not exist?

It says so and shows why: the one-sided limits disagree, or the function grows without bound. "Does not exist" and "infinite" are reported differently.

Does it handle limits at infinity?

Yes, including horizontal asymptotes and the standard comparisons between polynomial, exponential and logarithmic growth.

Can it prove a limit with epsilon-delta?

It will attempt one, and this is the weakest part of any AI solver. Treat a proof as a draft to check rather than a final answer.