Limits of Exponential Functions

The limit of the exponential function is an important formula in limit theory and is frequently used in limit computation.

Basic Form

Limit of the exponential function

lim⁡x→0ex−1x=1\lim_{x \to 0} \frac{e^x - 1}{x} = 1

This exponential limit is derived from the second important limit:

  1. The second important limit: lim⁡x→∞(1+1x)x=e\lim_{x \to \infty} \left(1 + \frac{1}{x}\right)^x = e
  2. Through a change of variable: lim⁡x→0(1+x)1x=e\lim_{x \to 0} (1 + x)^{\frac{1}{x}} = e
  3. Taking logarithms on both sides: lim⁡x→0ln⁡(1+x)x=1\lim_{x \to 0} \frac{\ln(1 + x)}{x} = 1

Detailed Derivation

Step 1: Change of variable

  • Let t=ex−1t = e^x - 1
  • Then ex=1+te^x = 1 + t
  • Taking logarithms on both sides: x=ln⁡(1+t)x = \ln(1 + t)

Step 2: Limit conversion As x→0x \to 0:

  • Since t=ex−1t = e^x - 1, as x→0x \to 0, ex→1e^x \to 1, so t→0t \to 0

Step 3: Establish the equality lim⁡x→0ex−1x=lim⁡t→0tln⁡(1+t)\lim_{x \to 0} \frac{e^x - 1}{x} = \lim_{t \to 0} \frac{t}{\ln(1 + t)}

Step 4: Use the second important limit The second important limit: lim⁡x→0ex−1x=1\lim_{x \to 0} \frac{e^x - 1}{x} = 1

Detailed explanation of the second important limit:

The original form of the second important limit is: lim⁡x→∞(1+1x)x=e\lim_{x \to \infty} \left(1 + \frac{1}{x}\right)^x = e

Through a change of variable, we get the reciprocal form: lim⁡x→0(1+x)1x=e\lim_{x \to 0} (1 + x)^{\frac{1}{x}} = e

Taking logarithms on both sides: lim⁡x→01xln⁡(1+x)=1\lim_{x \to 0} \frac{1}{x} \ln(1 + x) = 1

That is: lim⁡x→0ln⁡(1+x)x=1\lim_{x \to 0} \frac{\ln(1 + x)}{x} = 1

Using the inverse relationship between the exponential and logarithmic functions, we finally get: lim⁡x→0ex−1x=1\lim_{x \to 0} \frac{e^x - 1}{x} = 1

Therefore: lim⁡t→0tln⁡(1+t)=1\lim_{t \to 0} \frac{t}{\ln(1 + t)} = 1

Key insight: The core idea of this equality is that the exponential function and the logarithmic function are inverse operations. Through a change of variable, the limit of the exponential function is transformed into the limit of the logarithmic function, and the result is finally obtained using the known second important limit.

Generalized Form

lim⁡x→0ekx−1x=k\lim_{x \to 0} \frac{e^{kx} - 1}{x} = k

Equivalent Infinitesimal

Using this important limit, we obtain an important equivalent infinitesimal:

  • As x→0x \to 0, ex−1∼xe^x - 1 \sim x

Worked Examples

Example 1

Find lim⁡x→0e2x−1x\lim_{x \to 0} \frac{e^{2x} - 1}{x}

Solution: lim⁡x→0e2x−1x=2\lim_{x \to 0} \frac{e^{2x} - 1}{x} = 2


Practice Problems

Exercise 1

Find the limit lim⁡x→0e3x−1x\lim_{x \to 0} \frac{e^{3x} - 1}{x}.

exponential limit

Approach: Use the generalized form of the exponential limit.

Steps:

  1. lim⁡x→0e3x−1x=3\lim_{x \to 0} \frac{e^{3x} - 1}{x} = 3

Answer: The limit is 3.

Exercise 2

Find the limit lim⁡x→0e5x−12x\lim_{x \to 0} \frac{e^{5x} - 1}{2x}.

exponential limit

Approach: Use the generalized form of the exponential limit and a change of variable.

Steps:

  1. lim⁡x→0e5x−12x=lim⁡x→052⋅e5x−15x=52⋅1=52\lim_{x \to 0} \frac{e^{5x} - 1}{2x} = \lim_{x \to 0} \frac{5}{2} \cdot \frac{e^{5x} - 1}{5x} = \frac{5}{2} \cdot 1 = \frac{5}{2}

Answer: The limit is 52\frac{5}{2}.


Summary

Symbols Used in This Article

SymbolTypePronunciation/ExplanationMeaning in This Article
eeMathematical symbolnatural constantThe base of the natural logarithm, approximately 2.71828
lim⁡\limMathematical symbollimitThe limit of a function or sequence
→\toMathematical symboltends toA variable tending to some value
ln⁡\lnMathematical symbolnatural logarithmThe logarithm with base ee

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