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Limit of (1+x)2−1x\frac{\left(1 + x\right)^{2} - 1}{x} as x→0x \to 0

The limit of (1+x)2−1x\frac{\left(1 + x\right)^{2} - 1}{x} as x→0x \to 0 is 22.

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Graph of ((1 + x)^2 - 1) / x near x = 0-1.5-1-0.50.511.50.511.522.533.5limit = 2f(x)
Graph of ((1 + x)^2 - 1) / x near x = 0. The open circle marks the point x = 0, where the function may be undefined; the curve still heads to 2.

How to work it out

  1. 1

    We want what ((1 + x)^2 - 1) / x approaches as x → 0.

    lim⁡x→0(1+x)2−1x\lim_{x \to 0} \frac{\left(1 + x\right)^{2} - 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)2−1x=lim⁡2(1+x)1\lim \frac{\left(1 + x\right)^{2} - 1}{x} = \lim \frac{2 \left(1 + x\right)}{1}
  4. 4

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

    21=2\frac{2}{1} = 2
  5. 5

    Method: the derivative of (1+x)^2 at 0 is 2.

  6. 6

    Result:

    lim⁡x→0(1+x)2−1x=2\lim_{x \to 0} \frac{\left(1 + x\right)^{2} - 1}{x} = 2
LaTeX source of the formulas on this page
\[ \lim_{x \to 0} \frac{\left(1 + x\right)^{2} - 1}{x} = 2 \]
\[ \lim_{x \to 0} \frac{\left(1 + x\right)^{2} - 1}{x} \]
\[ \lim \frac{\left(1 + x\right)^{2} - 1}{x} = \lim \frac{2 \left(1 + x\right)}{1} \]
\[ \frac{2}{1} = 2 \]

Values of ((1 + x)^2 - 1) / x as x → 0 (computed)

xf(x)Distance from the limit
0.12.10000001.00e-1
-0.11.90000001.00e-1
0.012.01000001.00e-2
-0.011.99000001.00e-2
0.0012.00100001.00e-3
-0.0011.99900001.00e-3
0.00012.00010001.00e-4
-0.00011.99990001.00e-4
0.000012.00001001.00e-5
-0.000011.99999001.00e-5

Questions

What is the limit of ((1 + x)^2 - 1) / x as x approaches 0?

The limit of (1+x)2−1x\frac{\left(1 + x\right)^{2} - 1}{x} as x→0x \to 0 is 22.

Which form is it?

0/0.

More well-known limits

    Limit of ((1 + x)^2 - 1) / x as x → 0: 2 | WordGPT