HexMathv1.0.3

Limit calculator with steps

Type or scan a limit. HexMath finds its value and shows the algebra or rule used at every step.

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What this page covers

  • Limits at a point, including 00\frac{0}{0} forms that need factoring or rationalizing first.
  • Limits at infinity of polynomial, rational and exponential expressions.
  • One-sided limits, and L'Hôpital's rule for 00\frac{0}{0} and ∞∞\frac{\infty}{\infty} forms.
  • Not covered here: limits of sequences (use Sequence convergence calculator) and derivatives as answers in their own right (use Derivative calculator with steps).

How to enter the problem

  • Type lim, then the variable and the point it approaches, such as x→2, then the function. Write ∞ for infinity.
  • Scan it. Fit the one problem in the frame.
  • Choose a photo. Several problems in one photo are solved as a list.
  • Then tap Solve.

Worked examples

Example 1

Problem

Find lim⁡x→3x2−9x−3\displaystyle \lim_{x\to 3} \frac{x^2 - 9}{x - 3}.

Answer

Verified

The answer is

66

Explanation

  1. Try direct substitution
    At x=3x = 3 you get 00\frac{0}{0}. That is not a value; the expression must be rewritten first.
  2. Factor the numerator
    x2−9=(x−3)(x+3)x^2 - 9 = (x-3)(x+3)
  3. Cancel the common factor
    For x≠3x \neq 3, (x−3)(x+3)x−3=x+3\frac{(x-3)(x+3)}{x-3} = x + 3. The limit only looks at xx near 3, never at 3 itself.
  4. Substitute
    lim⁡x→3(x+3)=6\lim_{x\to 3} (x + 3) = 6

Example 2

Problem

Find lim⁡x→∞4x2+12x2−5x\displaystyle \lim_{x\to\infty} \frac{4x^2 + 1}{2x^2 - 5x}.

Answer

Verified

The answer is

22

Explanation

  1. Find the highest power
    The highest power of xx in the denominator is x2x^2.
  2. Divide top and bottom by x2x^2
    4+1/x22−5/x\frac{4 + 1/x^2}{2 - 5/x}.
  3. Let xx grow
    1x2→0\frac{1}{x^2} \to 0 and 5x→0\frac{5}{x} \to 0, so the expression approaches 42\frac{4}{2}.

Example 3

Problem

Find lim⁡x→0ex−1−xx2\displaystyle \lim_{x\to 0} \frac{e^x - 1 - x}{x^2}.

Answer

Verified

The answer is

12\frac{1}{2}

Explanation

  1. Check the form
    At x=0x = 0 the top is 1−1−0=01 - 1 - 0 = 0 and the bottom is 0, a 00\frac{0}{0} form, so L'Hôpital's rule applies.
  2. Apply L'Hôpital's rule once
    Differentiate the top and the bottom separately: ex−12x\frac{e^x - 1}{2x}. At x=0x = 0 this is still 00\frac{0}{0}.
  3. Apply it again
    ex2\frac{e^x}{2}, which is continuous at 0.
  4. Substitute
    e02=12\frac{e^0}{2} = \frac{1}{2}

Common mistakes

  • Writing 00\frac{0}{0}, or 0, as the answer. The form 00\frac{0}{0} only says the expression must be rewritten before the limit can be found.
  • At infinity, comparing the wrong terms. lim⁡x→∞4x2+12x2−5x\lim_{x\to\infty} \frac{4x^2+1}{2x^2-5x} is 2, the ratio of the x2x^2 coefficients, not −45-\frac{4}{5}.
  • Using the quotient rule inside L'Hôpital's rule. Differentiate the numerator and the denominator separately.
  • Using L'Hôpital's rule when the form is not 00\frac{0}{0} or ∞∞\frac{\infty}{\infty}. For x+1x\frac{x+1}{x} at x=1x = 1, substitution gives 2, while the rule would wrongly give 1.

Checks, assumptions and limits

  • Check numerically. For the runnable example, sin⁡(0.003)0.001≈2.9999955\frac{\sin(0.003)}{0.001} \approx 2.9999955, close to 3.
  • Angles are in radians. In degrees, sin⁡(3x)x\frac{\sin(3x)}{x} would approach π60\frac{\pi}{60} instead.
  • A two-sided limit exists only when the left and right limits agree. 1x\frac{1}{x} has different one-sided limits at 0, so its limit there does not exist.
  • HexMath can make mistakes. Double check important steps.

Frequently asked questions

The function does not settle on one finite value. The two one-sided limits differ, the values grow without bound, or they keep oscillating, as sin⁡(1/x)\sin(1/x) does near 0.

Only when direct substitution gives 00\frac{0}{0} or ∞∞\frac{\infty}{\infty}. Then the limit of fg\frac{f}{g} equals the limit of f′g′\frac{f'}{g'}, if that limit exists. Other forms, such as 0⋅∞0 \cdot \infty, must be rewritten as a fraction first.

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Last updated: · HexMath