Worked examples · Sheet 3

Standard limits and comparison of infinities

Theory used: Standard limits Comparison of infinities Rules for computing limits

Exercise 1

Compute

x \to 0 lim \frac{ln ( 1 + 3 x )}{x} and x \to 0 lim \frac{e ^{2 x} - 1}{5 x} .

\frac{ln ( 1 + 3 x )}{x} = 3 \cdot \frac{ln ( 1 + 3 x )}{3 x}, t = 3 x \to 0 : 3 \cdot \frac{ln ( 1 + t )}{t} \to 3

\frac{e ^{2 x} - 1}{5 x} = \frac{2}{5} \cdot \frac{e ^{2 x} - 1}{2 x}, t = 2 x \to 0 : \frac{2}{5} \cdot \frac{e ^{t} - 1}{t} \to \frac{2}{5}

Both limits are in the form $\frac{0}{0}$ and resemble the standard limits of the logarithm and of the exponential, but the argument is $3 x$ or $2 x$ instead of $x$ . We multiply and divide by the missing number, so that the same argument appears in the denominator, that is $3 x$ in the first case, $2 x$ in the second. With the substitution $t = 3 x$ , or $t = 2 x$ , the variable $t$ tends to $0$ when $x$ tends to $0$ , and the fraction becomes exactly a standard limit, which equals $1$ . The numerical factors remain, so the limits are $3$ and $\frac{2}{5}$ .

Exercise 2

Compute

x \to + \infty lim (1 + \frac{2}{x})^{x} and x \to + \infty lim (1 + \frac{1}{2 x})^{x} .

t = \frac{x}{2} : (1 + \frac{2}{x})^{x} = [(1 + \frac{1}{t})^{t}]^{2} \to e^{2}

t = 2 x : (1 + \frac{1}{2 x})^{x} = [(1 + \frac{1}{t})^{t}]^{\frac{1}{2}} \to e^{\frac{1}{2}} = e

The base tends to $1$ and the exponent to $+ \infty$ , so the form is $1^{\infty}$ , which is indeterminate. We try to reduce it to the standard limit $(1 + \frac{1}{t})^{t} \to e$ . In the first case we set $t = \frac{x}{2}$ , so that $\frac{2}{x} = \frac{1}{t}$ and the exponent $x$ becomes $2 t$ ; since $t \to + \infty$ , the power $(1 + \frac{1}{t})^{t}$ tends to $e$ and the limit is $e^{2}$ . In the second case we set $t = 2 x$ and the exponent $x$ becomes $\frac{t}{2}$ , so the limit is $e$ . The same procedure, with $\frac{r}{x}$ in place of $\frac{2}{x}$ , gives the limit $e^{r}$ of continuous compounding.

Exercise 3

Compute

x \to + \infty lim \frac{x ^{5} + e ^{x}}{2 e ^{x} - x ^{2}} and x \to + \infty lim \frac{ln x}{x} .

\frac{x ^{5} + e ^{x}}{2 e ^{x} - x ^{2}} = \frac{e ^{x} ( \frac{x ^{5}}{e ^{x}} + 1 )}{e ^{x} ( 2 - \frac{x ^{2}}{e ^{x}} )} = \frac{\frac{x ^{5}}{e ^{x}} + 1}{2 - \frac{x ^{2}}{e ^{x}}} \to \frac{0 + 1}{2 - 0} = \frac{1}{2}

\frac{ln x}{x} = \frac{ln x}{x ^{\frac{1}{2}}} \to 0

In the first quotient powers and exponentials appear. By the comparison of infinities the exponential $e^{x}$ grows faster than any power, so we factor out $e^{x}$ in the numerator and in the denominator. The ratios $\frac{x ^{5}}{e ^{x}}$ and $\frac{x ^{2}}{e ^{x}}$ tend to $0$ , and the limit is $\frac{1}{2}$ , because for large $x$ the powers are negligible compared with the exponentials. In the second quotient the logarithm is compared with the power $x^{\frac{1}{2}}$ , which grows faster than $ln x$ even though the exponent $\frac{1}{2}$ is small, therefore the limit is $0$ .