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Limit of (1+x)6−1x\frac{\left(1 + x\right)^{6} - 1}{x} as x→0x \to 0

The limit of (1+x)6−1x\frac{\left(1 + x\right)^{6} - 1}{x} as x→0x \to 0 is 66.

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Graph of ((1 + x)^6 - 1) / x near x = 0-1.5-1-0.50.511.520406080100120140limit = 6f(x)
Graph of ((1 + x)^6 - 1) / x near x = 0. The open circle marks the point x = 0, where the function may be undefined; the curve still heads to 6.

How to work it out

  1. 1

    We want what ((1 + x)^6 - 1) / x approaches as x → 0.

    lim⁡x→0(1+x)6−1x\lim_{x \to 0} \frac{\left(1 + x\right)^{6} - 1}{x}
  2. 2

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

  3. 3

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

    lim⁡(1+x)6−1x=lim⁡6(1+x)51\lim \frac{\left(1 + x\right)^{6} - 1}{x} = \lim \frac{6 \left(1 + x\right)^{5}}{1}
  4. 4

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

    61=6\frac{6}{1} = 6
  5. 5

    Method: the derivative of (1+x)^6 at 0 is 6.

  6. 6

    Result:

    lim⁡x→0(1+x)6−1x=6\lim_{x \to 0} \frac{\left(1 + x\right)^{6} - 1}{x} = 6
LaTeX source of the formulas on this page
\[ \lim_{x \to 0} \frac{\left(1 + x\right)^{6} - 1}{x} = 6 \]
\[ \lim_{x \to 0} \frac{\left(1 + x\right)^{6} - 1}{x} \]
\[ \lim \frac{\left(1 + x\right)^{6} - 1}{x} = \lim \frac{6 \left(1 + x\right)^{5}}{1} \]
\[ \frac{6}{1} = 6 \]

Values of ((1 + x)^6 - 1) / x as x → 0 (computed)

xf(x)Distance from the limit
0.17.71561001.72e+0
-0.14.68559001.31e+0
0.016.15201511.52e-1
-0.015.85198511.48e-1
0.0016.01502001.50e-2
-0.0015.98502001.50e-2
0.00016.00150021.50e-3
-0.00015.99850021.50e-3
0.000016.00015001.50e-4
-0.000015.99985001.50e-4

Questions

What is the limit of ((1 + x)^6 - 1) / x as x approaches 0?

The limit of (1+x)6−1x\frac{\left(1 + x\right)^{6} - 1}{x} as x→0x \to 0 is 66.

Which form is it?

0/0.

More well-known limits

    Limit of ((1 + x)^6 - 1) / x as x → 0: 6 | WordGPT