Exponential & Logarithmic Limits
In this reading, we study a powerful group of tools in calculus: exponential and logarithmic limits in calculus. These limits are central to differentiation and integration of functions like ##e^x## and ##\ln x##, and they appear in models of growth and decay, compound interest, and many examination-style questions.
Two fundamental limits will guide almost everything we do:
- ###\displaystyle \lim_{x \to 0} \frac{e^x – 1}{x} = 1###
- ###\displaystyle \lim_{x \to 0} \frac{\ln(1 + x)}{x} = 1###
Once we are comfortable with these, many other exponential and logarithmic limits in calculus become straightforward.
1. Why exponential and logarithmic limits matter
For trigonometric functions, the limit ###\lim_{x \to 0} \frac{\sin x}{x} = 1### plays a key role in differentiation. In exactly the same spirit, exponential and logarithmic limits in calculus provide the foundation for derivatives such as
- ###\dfrac{d}{dx}(e^x) = e^x,###
- ###\dfrac{d}{dx}(\ln x) = \dfrac{1}{x}.###
Behind these derivative formulas, the real engine is the pair of fundamental limits above.
2. The key exponential limit: \(\displaystyle \lim_{x\to 0} \frac{e^x – 1}{x} = 1\)
The first major result among exponential and logarithmic limits in calculus is
###\lim_{x \to 0} \frac{e^x – 1}{x} = 1.###
This means that near ##x = 0##, the graph of ##e^x## behaves almost like the straight line ##y = 1 + x##. Informally,
###e^x \approx 1 + x \quad \text{for small } x.###
One way to see this is through the power-series expansion of ##e^x##:
###e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots.###
We compute
###\frac{e^x – 1}{x}
= \frac{1 + x + \dfrac{x^2}{2!} + \dfrac{x^3}{3!} + \cdots – 1}{x}
= \frac{x + \dfrac{x^2}{2!} + \dfrac{x^3}{3!} + \cdots}{x}
= 1 + \frac{x}{2!} + \frac{x^2}{3!} + \cdots.###
As ##x \to 0##, all terms containing ##x, x^2, \dots## vanish, and we are left with
###\lim_{x \to 0} \frac{e^x – 1}{x} = 1.###
3. Scaling the exponential limit: \(\displaystyle \lim_{x\to 0} \frac{e^{kx} – 1}{x}\)
Let ##k## be a constant. Consider
###\lim_{x \to 0} \frac{e^{kx} – 1}{x}.###
We write
###\frac{e^{kx} – 1}{x} = k \cdot \frac{e^{kx} – 1}{kx}.###
Set ##t = kx##; then as ##x \to 0##, ##t \to 0## and
###\frac{e^{kx} – 1}{kx} = \frac{e^t – 1}{t}.###
Using the fundamental limit, ###\lim_{t \to 0} \frac{e^t – 1}{t} = 1###, so
###\lim_{x \to 0} \frac{e^{kx} – 1}{x} = k.###
4. The key logarithmic limit: \(\displaystyle \lim_{x\to 0} \frac{\ln(1+x)}{x} = 1\)
The second pillar of exponential and logarithmic limits in calculus is
###\lim_{x \to 0} \frac{\ln(1 + x)}{x} = 1.###
Near ##x = 0##, this tells us that ##\ln(1 + x)## behaves like the line ##y = x##:
###\ln(1 + x) \approx x \quad \text{for small } x.###
The power-series expansion of the natural logarithm near zero is
###\ln(1 + x) = x – \frac{x^2}{2} + \frac{x^3}{3} – \frac{x^4}{4} + \cdots.###
Then
###\frac{\ln(1 + x)}{x}
= \frac{x – \dfrac{x^2}{2} + \dfrac{x^3}{3} – \cdots}{x}
= 1 – \frac{x}{2} + \frac{x^2}{3} – \cdots.###
As ##x \to 0##, the terms involving ##x, x^2, \dots## vanish, giving
###\lim_{x \to 0} \frac{\ln(1 + x)}{x} = 1.###
5. Variations: \(\displaystyle \lim_{x \to 0} \frac{\ln(1 + a x)}{x}\)
Let ##a## be a constant. Consider
###\lim_{x \to 0} \frac{\ln(1 + a x)}{x}.###
We write
###\frac{\ln(1 + ax)}{x} = a \cdot \frac{\ln(1 + ax)}{ax}.###
Set ##t = ax##; as ##x \to 0##, ##t \to 0## and
###\frac{\ln(1 + t)}{t} \to 1.###
Thus
###\lim_{x \to 0} \frac{\ln(1 + ax)}{x} = a.###
6. Expansion-based thinking
For many exponential and logarithmic limits in calculus, it is convenient to remember simple approximations near ##x = 0##:
- ###e^x \approx 1 + x + \frac{x^2}{2}###
- ###\ln(1 + x) \approx x – \frac{x^2}{2}###
These allow us to identify the leading term in a numerator or denominator and decide whether a limit is finite, zero, or infinite.
7. Worked examples with exponential limits
Example 1
Evaluate ###\displaystyle \lim_{x \to 0} \frac{e^{2x} – 1}{x}.###
Solution.
Using the scaled limit with ##k = 2##,
###\lim_{x \to 0} \frac{e^{2x} – 1}{x} = 2.###
Example 2
Evaluate ###\displaystyle \lim_{x \to 0} \frac{e^{3x} – 1}{\ln(1 + 2x)}.###
Solution.
Write
###\frac{e^{3x} – 1}{\ln(1 + 2x)}
= \frac{\dfrac{e^{3x} – 1}{x}}{\dfrac{\ln(1 + 2x)}{x}}.###
As ##x \to 0##,
- ###\dfrac{e^{3x} – 1}{x} \to 3,###
- ###\dfrac{\ln(1 + 2x)}{x} \to 2.###
Therefore,
###\lim_{x \to 0} \frac{e^{3x} – 1}{\ln(1 + 2x)} = \frac{3}{2}.###
Example 3
Evaluate ###\displaystyle \lim_{x \to 0} \frac{e^x – 1 – x}{x^2}.###
Solution.
Using the expansion
###e^x = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \cdots,###
we get
###e^x – 1 – x = \frac{x^2}{2} + \frac{x^3}{6} + \cdots.###
Thus
###\frac{e^x – 1 – x}{x^2} = \frac{\frac{x^2}{2} + \frac{x^3}{6} + \cdots}{x^2}
= \frac{1}{2} + \frac{x}{6} + \cdots.###
As ##x \to 0##, this tends to ###\frac{1}{2}###.
Example 4
Evaluate ###\displaystyle \lim_{x \to 0} \frac{e^{kx} – 1 – kx}{x^2}.###
Solution.
Using the expansion
###e^{kx} = 1 + kx + \frac{k^2 x^2}{2} + \cdots,###
we find
###e^{kx} – 1 – kx = \frac{k^2 x^2}{2} + \cdots.###
Therefore,
###\frac{e^{kx} – 1 – kx}{x^2} \to \frac{k^2}{2}### as ##x \to 0##.
8. Worked examples with logarithmic limits
Example 5
Evaluate ###\displaystyle \lim_{x \to 0} \frac{\ln(1 + 3x)}{x}.###
Solution.
Using the scaled logarithmic limit with ##a = 3##,
###\lim_{x \to 0} \frac{\ln(1 + 3x)}{x} = 3.###
Example 6
Evaluate ###\displaystyle \lim_{x \to 0} \frac{\ln(1 + x) – x}{x^2}.###
Solution.
Using the expansion
###\ln(1 + x) = x – \frac{x^2}{2} + \frac{x^3}{3} – \cdots,###
we obtain
###\ln(1 + x) – x = -\frac{x^2}{2} + \frac{x^3}{3} – \cdots.###
Thus
###\frac{\ln(1 + x) – x}{x^2}
= -\frac{1}{2} + \frac{x}{3} – \cdots,###
and the limit is
###\lim_{x \to 0} \frac{\ln(1 + x) – x}{x^2} = -\frac{1}{2}.###
Example 7
Evaluate ###\displaystyle \lim_{x \to 0} \frac{\ln(1 + ax) – ax}{x^2}.###
Solution.
Using the expansion for ##\ln(1 + ax)##,
###\ln(1 + ax) = ax – \frac{a^2 x^2}{2} + \cdots,###
we obtain
###\ln(1 + ax) – ax = -\frac{a^2 x^2}{2} + \cdots.###
Therefore,
###\lim_{x \to 0} \frac{\ln(1 + ax) – ax}{x^2} = -\frac{a^2}{2}.###
Example 8
Evaluate ###\displaystyle \lim_{x \to 0} \left(1 + 2x\right)^{1/x}.###
Solution.
Let
###L = \lim_{x \to 0} (1 + 2x)^{1/x}.###
Taking natural logarithms,
###\ln L = \lim_{x \to 0} \frac{1}{x} \ln(1 + 2x).###
Write this as
###\ln L = \lim_{x \to 0} 2 \cdot \frac{\ln(1 + 2x)}{2x}.###
Using ###\lim_{u \to 0} \frac{\ln(1 + u)}{u} = 1### with ##u = 2x##, we get ###\frac{\ln(1 + 2x)}{2x} \to 1###. So
###\ln L = 2, \quad L = e^2.###
9. Mixed examples using exponential and logarithmic limits
Example 9
Evaluate ###\displaystyle \lim_{x \to 0} \frac{e^{3x} – 1 – 3x}{x^2}.###
Solution.
Using ###e^{3x} = 1 + 3x + \frac{9x^2}{2} + \cdots###, we obtain
###e^{3x} – 1 – 3x = \frac{9x^2}{2} + \cdots.###
Thus
###\frac{e^{3x} – 1 – 3x}{x^2} \to \frac{9}{2}.###
Example 10
Evaluate ###\displaystyle \lim_{x \to 0} \frac{e^x – \ln(1 + x) – 1}{x}.###
Solution.
Using expansions,
###e^x = 1 + x + \frac{x^2}{2} + \cdots,###
###\ln(1 + x) = x – \frac{x^2}{2} + \cdots.###
Then
###e^x – \ln(1 + x) – 1
= \left(1 + x + \frac{x^2}{2} + \cdots\right)
– \left(x – \frac{x^2}{2} + \cdots\right) – 1
= x + \frac{x^2}{2} + \cdots – x + \frac{x^2}{2} – \cdots
= x^2 + \cdots.###
Hence
###\frac{e^x – \ln(1 + x) – 1}{x} \approx \frac{x^2}{x} = x \to 0.###
So the limit is 0.
Example 11
Evaluate ###\displaystyle \lim_{x \to 0} \frac{\ln(1 + x)}{e^x – 1}.###
Solution.
We write
###\frac{\ln(1 + x)}{e^x – 1}
= \frac{\ln(1 + x)/x}{(e^x – 1)/x}.###
As ##x \to 0##, both the numerator and denominator tend to 1, so
###\lim_{x \to 0} \frac{\ln(1 + x)}{e^x – 1} = 1.###
Example 12
Evaluate ###\displaystyle \lim_{x \to 0} \left(\frac{1 + x}{1 – x}\right)^{1/x}.###
Solution.
Let
###L = \lim_{x \to 0} \left(\frac{1 + x}{1 – x}\right)^{1/x}.###
Take natural logarithms:
###\ln L = \lim_{x \to 0} \frac{1}{x} \ln \left(\frac{1 + x}{1 – x}\right)
= \lim_{x \to 0} \frac{\ln(1 + x) – \ln(1 – x)}{x}.###
Split the fraction:
###\ln L = \lim_{x \to 0} \left[\frac{\ln(1 + x)}{x} – \frac{\ln(1 – x)}{x}\right].###
We know
- ###\displaystyle \frac{\ln(1 + x)}{x} \to 1,###
- For the second term, write ###\frac{\ln(1 – x)}{x} = -\frac{\ln(1 – x)}{-x}###, and with ##t = -x##, ###\frac{\ln(1 + t)}{t} \to 1###, so ###\frac{\ln(1 – x)}{x} \to -1.###
Thus
###\ln L = 1 – (-1) = 2.###
Hence ###L = e^2.###
10. References for further reading
- Background on the exponential function and the constant ##e## can be found in the article on the exponential function at Wikipedia.
- The article on the natural logarithm at Wikipedia provides further examples and graphs illustrating logarithmic behaviour.
- Research and expository notes from organisations such as the American Mathematical Society (AMS) and the National Institute of Standards and Technology (NIST) often discuss exponential and logarithmic functions in analytic contexts.
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