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Math III

Limits of functions

Find limits as x→ax \to a or x→∞x \to \infty. Learn how to transform indeterminate forms and a few essential limits.

Basic, Standard: Grade 12 · Term 1 / Advanced: Grade 12 · Exam prep

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Math problem generator

Level

0/0 forms: factoring and rationalizing

If substituting x=ax = a gives 00\frac{0}{0}, the numerator and denominator share the factor x−ax - a. Factor and cancel before substituting.

lim⁡x→2x2−5x+6x−2=lim⁡x→2(x−3)=−1\lim_{x \to 2} \frac{x^2 - 5x + 6}{x - 2} = \lim_{x \to 2} (x - 3) = -1

With square roots, rationalize:

lim⁡x→0x+9−3x=lim⁡x→01x+9+3=16\lim_{x \to 0} \frac{\sqrt{x + 9} - 3}{x} = \lim_{x \to 0} \frac{1}{\sqrt{x + 9} + 3} = \frac{1}{6}
Unknown coefficients

If lim⁡x→af(x)x−a\displaystyle\lim_{x \to a} \frac{f(x)}{x - a} is finite, the denominator tends to 00, so the numerator must satisfy f(a)=0f(a) = 0 (a necessary condition). Use this to find unknown constants.

Trigonometric limits

Formulas
lim⁡x→0sin⁡xx=1,lim⁡x→01−cos⁡xx2=12\lim_{x \to 0} \frac{\sin x}{x} = 1,\qquad \lim_{x \to 0} \frac{1 - \cos x}{x^2} = \frac{1}{2}

Create the form sin⁡θθ\frac{\sin \theta}{\theta} first, as in sin⁡3x2x=sin⁡3x3x⋅32\frac{\sin 3x}{2x} = \frac{\sin 3x}{3x} \cdot \frac{3}{2}. Multiply expressions with 1−cos⁡x1 - \cos x by 1+cos⁡x1 + \cos x to get sin⁡2x\sin^2 x.

For limits such as x→πx \to \pi, substitute t=x−πt = x - \pi to get a limit as t→0t \to 0.

The number $e$ and exponential limits

Formulas
lim⁡h→0(1+h)1h=e,lim⁡x→±∞(1+1x)x=e\lim_{h \to 0} (1 + h)^{\frac{1}{h}} = e,\qquad \lim_{x \to \pm\infty} \left(1 + \frac{1}{x}\right)^x = e
lim⁡h→0eh−1h=1,lim⁡h→0log⁡(1+h)h=1\lim_{h \to 0} \frac{e^h - 1}{h} = 1,\qquad \lim_{h \to 0} \frac{\log(1 + h)}{h} = 1

Example: lim⁡x→0(1+2x)1x=lim⁡x→0{(1+2x)12x}2=e2\displaystyle\lim_{x \to 0} (1 + 2x)^{\frac{1}{x}} = \lim_{x \to 0} \left\{(1 + 2x)^{\frac{1}{2x}}\right\}^2 = e^2. Adjust the exponent to match the definition. (Here log⁡\log is the natural logarithm.)

One-sided limits and continuity

The right-hand limit lim⁡x→a+f(x)\displaystyle\lim_{x \to a^{+}} f(x) is taken as xx approaches aa from above, and the left-hand limit lim⁡x→a−f(x)\displaystyle\lim_{x \to a^{-}} f(x) from below. The limit lim⁡x→af(x)\displaystyle\lim_{x \to a} f(x) exists exactly when they agree.

ff is continuous at x=ax = a when lim⁡x→af(x)=f(a)\displaystyle\lim_{x \to a} f(x) = f(a). Remove absolute values like ∣x−a∣|x - a| by treating x>ax > a and x<ax < a separately.