calculate the limit of x->0 (sqrt(x+1)-1)/x

1) Compute the limit of

\frac{\sqrt{x + 1} - 1}{x}

x \to 0

ⓘ The limit of a sum of finite limits is the sum of the limits

The limit of

x + 1

1

when

x \to 0

ⓘ The function sqrt is continuous, its limit is its value at the limit of its argument

The limit of

\sqrt{x + 1}

\sqrt{1}

when

x \to 0

⇥1.1) Let's calculate:

\sqrt{1}

⇥

1

ⓘ The limit of a sum of finite limits is the sum of the limits

The limit of

\sqrt{x + 1} - 1

1 - 1

when

x \to 0

⇥1.2) Let's calculate:

1 - 1

⇥ⓘ Start from

1

and move back

1

on the number line, down to zero:

1 - 1

0

⇥

0

For

\frac{\sqrt{x + 1} - 1}{x}

, we encounter the indeterminate form 0/0 as

x

tends to

0

2) We look for a Taylor expansion of the expression:

\sqrt{x + 1}

Using Taylor's formula:

\sqrt{A} + \frac{X - A}{2 \sqrt{A}}

for the function sqrt, we compute the Taylor expansion for X=

x + 1

at the point A=

1

of order N=1

\sqrt{1} + \frac{x + 1 - 1}{2 \cdot \sqrt{1}}

ⓘ The numerator and the denominator are computed before the division:

1 - 1

0

1 + \frac{x + 0}{2}

3) We redo the calculation using the Taylor expansion:

1 + \frac{x}{2}

\frac{(1 + \frac{x}{2}) - 1}{x}

ⓘ Calculate the numerator and denominator before dividing:

1 - 1

0

■Add fractions by using a common denominator and then adding their numerators:

0 + \frac{x}{2}

\frac{0 \cdot 2 + x}{2}

\frac{\frac{0 \cdot 2 + x}{2}}{x}

ⓘ If a factor is zero, the product is zero:

0 \cdot 2

0

\frac{\frac{0 + x}{2}}{x}

ⓘ Adding 0 changes nothing:

0 + x

x

\frac{\frac{x}{2}}{x}

ⓘ Dividing a fraction by

x

is multiplying its denominator by

x

\frac{\frac{x}{2}}{x}

\frac{x}{2 x}

\frac{x}{2 x}

ⓘ The original denominator must remain nonzero. The fraction is simplified by dividing the numerator and the denominator by their common factor

x

\frac{x}{2 x}

\frac{1}{2}

\frac{1}{2}

So the limit of

\frac{\sqrt{x + 1} - 1}{x}

\frac{1}{2}

when

x \to 0

▶Limit:

\frac{1}{2}

4) The numeric value of:

\frac{1}{2}

is:

▷

0.5