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Limit of 3xβˆ’1x\frac{3^{x} - 1}{x} as xβ†’0x \to 0

The limit of 3xβˆ’1x\frac{3^{x} - 1}{x} as xβ†’0x \to 0 is ln⁑3\ln 3.

Graph of (3^x - 1) / x near x = 0-1.5-1-0.50.511.50.511.522.5limit = ln 3 β‰ˆ 1.0986123f(x)
Graph of (3^x - 1) / x near x = 0. The open circle marks the point x = 0, where the function may be undefined; the curve still heads to ln 3 β‰ˆ 1.0986123.

How to work it out

  1. 1

    We want what (3x3^{x} - 1) / x approaches as x β†’ 0.

    lim⁑xβ†’03xβˆ’1x\lim_{x \to 0} \frac{3^{x} - 1}{x}
  2. 2

    Direct substitution gives an indeterminate form: 0/0. It does not tell us the limit.

  3. 3

    L'HΓ΄pital's rule: when the form is 0/0, the limit equals the limit of the ratio of the derivatives.

    lim⁑3xβˆ’1x=lim⁑3xln⁑(3)1\lim \frac{3^{x} - 1}{x} = \lim \frac{3^{x} \ln\left(3\right)}{1}
  4. 4

    Now substitute x = 0: the denominator is not 0 any more.

    1.09861231=1.0986123\frac{1.0986123}{1} = 1.0986123
  5. 5

    Method: the derivative of 3x3^{x} at 0 is ln 3.

  6. 6

    Result:

    lim⁑xβ†’03xβˆ’1x=ln⁑3\lim_{x \to 0} \frac{3^{x} - 1}{x} = \ln 3
LaTeX source of the formulas on this page
\[ \lim_{x \to 0} \frac{3^{x} - 1}{x} = \ln 3 \]
\[ \lim_{x \to 0} \frac{3^{x} - 1}{x} \]
\[ \lim \frac{3^{x} - 1}{x} = \lim \frac{3^{x} \ln\left(3\right)}{1} \]
\[ \frac{1.0986123}{1} = 1.0986123 \]

Values of (3x3^{x} - 1) / x as x β†’ 0 (computed)

xf(x)Distance from the limit
0.11.16123176.26e-2
-0.11.04041545.82e-2
0.011.10466926.06e-3
-0.011.09259966.01e-3
0.0011.09921606.04e-4
-0.0011.09800906.03e-4
0.00011.09867266.03e-5
-0.00011.09855196.03e-5
0.000011.09861836.03e-6
-0.000011.09860636.03e-6

Questions

What is the limit of (3x3^{x} - 1) / x as x approaches 0?

The limit of 3xβˆ’1x\frac{3^{x} - 1}{x} as xβ†’0x \to 0 is ln⁑3\ln 3.

Which form is it?

0/0.

More well-known limits

    Limit of (3^x - 1) / x as x β†’ 0: ln 3 β‰ˆ 1.0986123 | WordGPT