Limit Calculator

Evaluate limits of functions with AI-powered step-by-step solutions

Drag & drop or click to add images or PDF

Math Input
limit of sin(x)/x as x -> 0
limit of (1 + 1/n)^n as n -> infinity
limit of (x^2 - 4)/(x - 2) as x -> 2
limit of x*ln(x) as x -> 0+

What is a Limit?

A limit describes the value that a function approaches as the input approaches a particular point. The formal definition states:

limxaf(x)=L\lim_{x \to a} f(x) = L

means that for every ϵ>0\epsilon > 0, there exists a δ>0\delta > 0 such that if 0<xa<δ0 < |x - a| < \delta, then f(x)L<ϵ|f(x) - L| < \epsilon.

Intuitively, a limit answers: "What value does f(x)f(x) get arbitrarily close to as xx gets close to aa?"

One-sided limits approach from a single direction:

  • Left-hand limit: limxaf(x)\lim_{x \to a^-} f(x)
  • Right-hand limit: limxa+f(x)\lim_{x \to a^+} f(x)

A two-sided limit exists only when both one-sided limits exist and are equal.

Limits at infinity describe end behavior:

limxf(x)=L\lim_{x \to \infty} f(x) = L

means f(x)f(x) approaches LL as xx grows without bound.

Limits are foundational to calculus — they define derivatives, integrals, and continuity. A function is continuous at aa if and only if limxaf(x)=f(a)\lim_{x \to a} f(x) = f(a).

How to Evaluate Limits

Method 1: Direct Substitution

The simplest approach — plug in the value. If f(a)f(a) is defined and the function is continuous at aa:

limxaf(x)=f(a)\lim_{x \to a} f(x) = f(a)

Example: limx3(x2+1)=9+1=10\lim_{x \to 3} (x^2 + 1) = 9 + 1 = 10

Method 2: Factoring and Cancellation

When direct substitution yields 00\frac{0}{0}, factor and cancel:

limx2x24x2=limx2(x2)(x+2)x2=limx2(x+2)=4\lim_{x \to 2} \frac{x^2 - 4}{x - 2} = \lim_{x \to 2} \frac{(x-2)(x+2)}{x-2} = \lim_{x \to 2}(x+2) = 4

Method 3: L'Hôpital's Rule

When direct substitution gives 00\frac{0}{0} or \frac{\infty}{\infty}:

limxaf(x)g(x)=limxaf(x)g(x)\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)}

provided the right-hand limit exists.

Example: limx0sinxx=limx0cosx1=1\lim_{x \to 0} \frac{\sin x}{x} = \lim_{x \to 0} \frac{\cos x}{1} = 1

Method 4: Squeeze Theorem

If g(x)f(x)h(x)g(x) \leq f(x) \leq h(x) near aa, and limxag(x)=limxah(x)=L\lim_{x \to a} g(x) = \lim_{x \to a} h(x) = L, then limxaf(x)=L\lim_{x \to a} f(x) = L.

Method 5: Conjugate Multiplication

For expressions with radicals:

limx0x+42x=limx0(x+42)(x+4+2)x(x+4+2)=limx0xx(x+4+2)=14\lim_{x \to 0} \frac{\sqrt{x+4} - 2}{x} = \lim_{x \to 0} \frac{(\sqrt{x+4}-2)(\sqrt{x+4}+2)}{x(\sqrt{x+4}+2)} = \lim_{x \to 0} \frac{x}{x(\sqrt{x+4}+2)} = \frac{1}{4}

Important Standard Limits

LimitValue
limx0sinxx\lim_{x \to 0} \frac{\sin x}{x}11
limx0ex1x\lim_{x \to 0} \frac{e^x - 1}{x}11
limx0ln(1+x)x\lim_{x \to 0} \frac{\ln(1+x)}{x}11
limn(1+1n)n\lim_{n \to \infty} (1 + \frac{1}{n})^nee
limx01cosxx2\lim_{x \to 0} \frac{1 - \cos x}{x^2}12\frac{1}{2}

Comparison of Methods

MethodBest ForKey Indicator
Direct SubstitutionContinuous functionsNo indeterminate form
FactoringPolynomial 00\frac{0}{0}Both num/denom have common factor
L'Hôpital's Rule00\frac{0}{0} or \frac{\infty}{\infty}Indeterminate quotient
Squeeze TheoremOscillating functionsBounded between known limits
ConjugateRadical expressions\sqrt{\cdot} in numerator/denominator

Common Mistakes to Avoid

  • Applying L'Hôpital's Rule without verifying indeterminate form: The rule only applies to 00\frac{0}{0} or \frac{\infty}{\infty}. Using it on 10\frac{1}{0} or other forms gives wrong answers.
  • Confusing limit existence with function value: limxaf(x)\lim_{x \to a} f(x) can exist even if f(a)f(a) is undefined. The limit depends on nearby values, not the value at the point.
  • Ignoring one-sided limits: For piecewise functions or at discontinuities, always check both left and right limits separately.
  • Incorrectly distributing limits over indeterminate arithmetic: lim(fg)limflimg\lim(f - g) \neq \lim f - \lim g when both are \infty (gives \infty - \infty, which is indeterminate).
  • Treating \frac{\infty}{\infty} as 1: \frac{\infty}{\infty} is indeterminate — it can equal any value.

Examples

Step 1: Direct substitution gives e010=00\frac{e^0 - 1}{0} = \frac{0}{0} (indeterminate)
Step 2: Apply L'Hôpital's Rule: differentiate numerator and denominator
Step 3: limx0ex1=e0=1\lim_{x \to 0} \frac{e^x}{1} = e^0 = 1
Answer: 11

Step 1: Both numerator and denominator approach \infty. Divide every term by x2x^2:
Step 2: limx3+2x51x2\lim_{x \to \infty} \frac{3 + \frac{2}{x}}{5 - \frac{1}{x^2}}
Step 3: As xx \to \infty: 2x0\frac{2}{x} \to 0 and 1x20\frac{1}{x^2} \to 0, so the limit equals 35\frac{3}{5}
Answer: 35\frac{3}{5}

Step 1: Direct substitution gives 00\frac{0}{0}. Rewrite using the standard limit limt0sintt=1\lim_{t \to 0} \frac{\sin t}{t} = 1:
Step 2: sin(3x)sin(5x)=sin(3x)3x5xsin(5x)3x5x\frac{\sin(3x)}{\sin(5x)} = \frac{\sin(3x)}{3x} \cdot \frac{5x}{\sin(5x)} \cdot \frac{3x}{5x}
Step 3: As x0x \to 0: each fraction involving sine approaches 1, leaving 35\frac{3}{5}
Answer: 35\frac{3}{5}

Frequently Asked Questions

An indeterminate form is an expression like 0/0, infinity/infinity, 0 times infinity, infinity minus infinity, 0^0, 1^infinity, or infinity^0. These forms do not have a predetermined value and require further analysis to evaluate.

You can use L'Hopital's Rule only when direct substitution gives the indeterminate form 0/0 or infinity/infinity. Both the numerator and denominator must be differentiable near the point, and the limit of the ratio of derivatives must exist.

Yes. The limit depends on what the function approaches near the point, not its value at the point. For example, (x^2 - 1)/(x - 1) is undefined at x = 1, but its limit as x approaches 1 is 2.

When a limit equals infinity, it means the function grows without bound as x approaches the given value. Technically the limit does not exist as a finite number, but we write the limit equals infinity to describe this specific unbounded behavior.

Related Solvers

DerivativeIntegralSeriesDifferential equation
Try AI-Math for Free

Get step-by-step solutions to any math problem. Upload a photo or type your question.

Start Solving