Limits as x→∞ (convergence and divergence)

Divide the numerator and denominator by the highest power in the denominator. If the denominator has the higher degree, the limit is $0$; if the degrees are equal, the limit is the ratio of the leading coefficients.

If the numerator has the higher degree, the limit diverges. For example, $\frac{x^3+5}{x^2+3}$ grows at about the same rate as $x$, so it diverges to positive infinity. Whether it goes to positive or negative infinity depends on the sign of $\frac{a}{b}$.

Examples

1. 01Which one diverges as x→∞?$\displaystyle$A$\displaystyle \frac{x+4}{x^{2}+4}$B$\displaystyle \frac{x^{2}+4}{x+4}$C$\displaystyle \frac{5x+1}{x+4}$D$\displaystyle \frac{x^{2}+4}{5x^{2}}$AnswerHide

   Answer$\displaystyle \frac{x^{2}+4}{x+4}$

   Steps

   1. $\displaystyle \frac{x^{2}+4}{x+4}$$\displaystyle = \frac{x+\frac{4}{x}}{1+\frac{4}{x}}$
   2. Here everything vanishes instead
   3. $\displaystyle \frac{x+4}{x^{2}+4}$$\displaystyle = \frac{\frac{1}{x}+\frac{4}{x^{2}}}{1+\frac{4}{x^{2}}}$
2. 02What is the value of this limit?$\displaystyle \lim_{x \to \infty} \frac{4x^{2}+x}{3x^{2}+2}$AnswerHide

   Answer$\displaystyle \frac{4}{3}$
3. 03What is the value of this limit?$\displaystyle \lim_{x \to \infty} \frac{2x+1}{5x^{2}+1}$AnswerHide

   Answer$\displaystyle 0$

   Steps

   1. $\displaystyle \frac{2x+1}{5x^{2}+1}$$\displaystyle = \frac{\frac{2}{x}+\frac{1}{x^{2}}}{5+\frac{1}{x^{2}}}$
   2. Everything on top approaches 0
   3. $\displaystyle \lim_{x \to \infty} \frac{2x+1}{5x^{2}+1}$$\displaystyle = \frac{0}{5} = 0$
4. 04What is the value of this limit?$\displaystyle \lim_{x \to \infty} \frac{4x+1}{2x^{2}+1}$AnswerHide

   Answer$\displaystyle 0$

   Steps

   1. $\displaystyle \frac{4x+1}{2x^{2}+1}$$\displaystyle = \frac{\frac{4}{x}+\frac{1}{x^{2}}}{2+\frac{1}{x^{2}}}$
   2. Everything on top approaches 0
   3. $\displaystyle \lim_{x \to \infty} \frac{4x+1}{2x^{2}+1}$$\displaystyle = \frac{0}{2} = 0$
5. 05What is the value of this limit?$\displaystyle \lim_{x \to \infty} \frac{5x+1}{4x^{2}+1}$AnswerHide

   Answer$\displaystyle 0$

   Steps

   1. $\displaystyle \frac{5x+1}{4x^{2}+1}$$\displaystyle = \frac{\frac{5}{x}+\frac{1}{x^{2}}}{4+\frac{1}{x^{2}}}$
   2. Everything on top approaches 0
   3. $\displaystyle \lim_{x \to \infty} \frac{5x+1}{4x^{2}+1}$$\displaystyle = \frac{0}{4} = 0$
6. 06What is the value of this limit?$\displaystyle \lim_{x \to \infty} \frac{2x+1}{3x^{2}+1}$AnswerHide

   Answer$\displaystyle 0$

   Steps

   1. $\displaystyle \frac{2x+1}{3x^{2}+1}$$\displaystyle = \frac{\frac{2}{x}+\frac{1}{x^{2}}}{3+\frac{1}{x^{2}}}$
   2. Everything on top approaches 0
   3. $\displaystyle \lim_{x \to \infty} \frac{2x+1}{3x^{2}+1}$$\displaystyle = \frac{0}{3} = 0$

FAQ

Q1Isn't the limit of $\frac{2x+1}{5x^2+1}$ equal to $\frac{2}{5}$?AnswerHide

A

The ratio of coefficients is the answer only when the degrees are equal. Here the numerator has degree 1 and the denominator degree 2. Dividing by $x^2$ gives $\frac{\frac{2}{x}+\frac{1}{x^2}}{5+\frac{1}{x^2}}$, which converges to $0$.