Mathematics

Limits of Functions

Calculation of limits, operations with infinity, indeterminate forms, and equivalent infinitesimals.

0k=0 \displaystyle \frac{0}{k} = 0
k0=±∞ (IND) \displaystyle \frac{k}{0} = \pm\infty \text{ (IND)}
k∞=0 \displaystyle \frac{k}{\infty} = 0
∞k=∞ \displaystyle \frac{\infty}{k} = \infty
0∞=0 \displaystyle \frac{0}{\infty} = 0
∞0=∞ \displaystyle \frac{\infty}{0} = \infty
00=IND \displaystyle \frac{0}{0} = IND
∞∞=IND \displaystyle \frac{\infty}{\infty} = IND
∞−∞=IND \displaystyle \infty - \infty = IND
k⋅∞=∞ \displaystyle k \cdot \infty = \infty
∞⋅∞=∞ \displaystyle \infty \cdot \infty = \infty
0⋅∞=IND \displaystyle 0 \cdot \infty = IND
0k={0if k>0∞if k<0 \displaystyle 0^k = \begin{cases} 0 & \text{if } k > 0 \\ \infty & \text{if } k < 0 \end{cases}
k∞={∞if k>10if 0<k<1 \displaystyle k^\infty = \begin{cases} \infty & \text{if } k > 1 \\ 0 & \text{if } 0 < k < 1 \end{cases}
0∞=0 \displaystyle 0^\infty = 0
00=IND \displaystyle 0^0 = IND
∞0=IND \displaystyle \infty^0 = IND
1∞=IND \displaystyle 1^\infty = IND
Limits of functions at infinity can be approximated by simpler functions by applying the following rules of dominance (orders of infinity):
  • Between two exponential functions with bases greater than 1, the one with the larger base dominates.
  • An exponential function with a base greater than 1 dominates any power function.
  • Between two power functions of xx, the one with the larger exponent dominates.
  • Power functions of xx dominate logarithmic functions.
  • Two polynomials of the same degree, or two exponential functions with the same base, are infinities of the same order.
xx≫kx≫xa≫log⁡xx^x \gg k^x \gg x^a \gg \log x
When evaluating limits as x→0x \to 0, the following approximations can be used:
lim⁡x→0x∼lim⁡x→0sin⁡x∼lim⁡x→0tan⁡x∼lim⁡x→0arcsin⁡x∼lim⁡x→0arctan⁡x\lim_{x\to 0} x \sim \lim_{x\to 0} \sin x \sim \lim_{x\to 0} \tan x \sim \lim_{x\to 0} \arcsin x \sim \lim_{x\to 0} \arctan x
lim⁡x→0ex∼lim⁡x→0(1+x)\lim_{x\to 0} e^x \sim \lim_{x\to 0} (1+x)
lim⁡x→0cos⁡x∼lim⁡x→0(1−x22)\lim_{x\to 0} \cos x \sim \lim_{x\to 0} \left(1 - \frac{x^2}{2}\right)
lim⁡x→1ln⁡x∼lim⁡x→1(x−1)∼lim⁡x→1sin⁡(x−1)\lim_{x\to 1} \ln x \sim \lim_{x\to 1} (x-1) \sim \lim_{x\to 1} \sin(x-1)
This is solved by finding the one-sided (lateral) limits. The solutions will be ±∞\pm\infty, corresponding to a vertical asymptote. Example:
lim⁡x→22x−2=20\lim_{x\to 2} \frac{2}{x-2} = \frac{2}{0}
We find the one-sided limits:
lim⁡x→2−2x−2=20−=−∞\displaystyle \lim_{x\to 2^-} \frac{2}{x-2} = \frac{2}{0^-} = -\infty
lim⁡x→2+2x−2=20+=+∞\displaystyle \lim_{x\to 2^+} \frac{2}{x-2} = \frac{2}{0^+} = +\infty
Graph of one-sided limits at a vertical asymptote
This is solved by factoring and simplifying. If radicals (roots) are present, multiply and divide by the conjugate to eliminate the radicals before factoring. Example:
lim⁡x→2x2−4x2−3x+2=00\lim_{x\to 2} \frac{x^2 - 4}{x^2 - 3x + 2} = \frac{0}{0}
We factor, simplify, and evaluate the limit:
lim⁡x→2(x+2)(x−2)(x−2)(x−1)=lim⁡x→2x+2x−1=4\lim_{x\to 2} \frac{(x+2)(x-2)}{(x-2)(x-1)} = \lim_{x\to 2} \frac{x+2}{x-1} = 4
Method 1: Divide the numerator and denominator by the highest power of xx.

CAREFUL! If polynomials are inside square roots, their degree is halved. Example:
lim⁡x→∞x4−1x2−2=∞∞\lim_{x\to \infty} \frac{\sqrt{x^4 - 1}}{x^2 - 2} = \frac{\infty}{\infty}
We solve by dividing by x2x^2:
lim⁡x→∞x4−1x2x2−2x2=lim⁡x→∞x4x4−1x4x2x2−2x2=1−01−0=1\lim_{x\to \infty} \frac{\frac{\sqrt{x^4 - 1}}{x^2}}{\frac{x^2 - 2}{x^2}} = \lim_{x\to \infty} \frac{\sqrt{\frac{x^4}{x^4} - \frac{1}{x^4}}}{\frac{x^2}{x^2} - \frac{2}{x^2}} = \frac{\sqrt{1-0}}{1-0} = 1
Method 2: Approximate each polynomial by discarding the lower-degree terms, then simplify. Example:
lim⁡x→∞x4−1x2−2≈lim⁡x→∞x4x2=lim⁡x→∞x2x2=1\lim_{x\to \infty} \frac{\sqrt{x^4 - 1}}{x^2 - 2} \approx \lim_{x\to \infty} \frac{\sqrt{x^4}}{x^2} = \lim_{x\to \infty} \frac{x^2}{x^2} = 1
Method 3: By comparing orders of infinity: see the previous section.
Method 1: If there are algebraic fractions, combine them (find a common denominator), resulting in a previous indeterminate form (∞/∞\infty/\infty or k/0k/0), which is solved accordingly.
  • If there are irrational expressions (radicals), multiply and divide by the conjugate. This will lead to one of the previous indeterminate forms.
Method 2: By comparing orders of infinity: see the previous section. The higher-order infinity dominates.
Apply the following formula, where the new exponent will result in one of the previously mentioned indeterminate forms.
lim⁡x→kf(x)g(x)=elim⁡x→kg(x)⋅[f(x)−1]\lim_{x\to k} f(x)^{g(x)} = e^{\lim_{x\to k} g(x) \cdot [f(x) - 1]}