limit as x approaches 0 of (sqrt(5+x)-sqrt(5))/(2x)

1) Compute the limit of

\frac{\sqrt{5 + x} - \sqrt{5}}{2 x}

x \to 0

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

The limit of

5 + x

5

when

x \to 0

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

The limit of

\sqrt{5 + x}

\sqrt{5}

when

x \to 0

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

The limit of

- \sqrt{5}

- \sqrt{5}

when

x \to 0

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

The limit of

\sqrt{5 + x} - \sqrt{5}

\sqrt{5} - \sqrt{5}

when

x \to 0

⇥1.1) Let's evaluate:

\sqrt{5} - \sqrt{5}

⇥ⓘ Like terms are grouped by adding their coefficients:

\sqrt{5} - \sqrt{5}

0

⇥

0

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

The limit of

2 x

2 \cdot 0

when

x \to 0

⇥1.2) Let's evaluate:

2 \cdot 0

⇥ⓘ A product with a zero factor is zero:

2 \cdot 0

0

⇥

0

For

\frac{\sqrt{5 + x} - \sqrt{5}}{2 x}

, the limit takes the indeterminate form 0/0 when

x

→

0

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

\sqrt{5 + x}

Using Taylor's formula:

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

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

5 + x

at the point A=

5

of order N=1

\sqrt{5} + \frac{5 + x - 5}{2 \cdot \sqrt{5}}

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

5 - 5

0

\sqrt{5} + \frac{0 + x}{2 \cdot \sqrt{5}}

ⓘ Adding 0 changes nothing:

0 + x

x

\sqrt{5} + \frac{x}{2 \cdot \sqrt{5}}

The conventional preference is to have numerical square roots in the numerator rather than in the denominator:

ⅰ

\frac{x}{2 \cdot \sqrt{5}}

\frac{\sqrt{5} \cdot x}{5 \cdot 2}

\sqrt{5} + \frac{\sqrt{5} \cdot x}{5 \cdot 2}

3) We redo the calculation using the Taylor expansion:

\sqrt{5} + \frac{\sqrt{5} \cdot x}{10}

\frac{(\sqrt{5} + \frac{\sqrt{5} \cdot x}{10}) - \sqrt{5}}{2 x}

ⓘ Like terms are grouped by adding their coefficients:

\sqrt{5} - \sqrt{5}

0

■To add fractions, put them over the same denominator, then add the numerators:

0 + \frac{\sqrt{5} \cdot x}{10}

\frac{0 \cdot 10 + \sqrt{5} \cdot x}{10}

\frac{\frac{0 \cdot 10 + \sqrt{5} \cdot x}{10}}{2 x}

ⓘ A product with a zero factor is zero:

0 \cdot 10

0

\frac{\frac{0 + \sqrt{5} \cdot x}{10}}{2 x}

ⓘ Adding 0 changes nothing:

0 + \sqrt{5} \cdot x

\sqrt{5} \cdot x

\frac{\frac{\sqrt{5} \cdot x}{10}}{2 x}

ⓘ Dividing a fraction by

2 x

is multiplying its denominator by

2 x

\frac{\frac{\sqrt{5} \cdot x}{10}}{2 x}

\frac{\sqrt{5} \cdot x}{10 \cdot 2 x}

\frac{\sqrt{5} \cdot x}{10 \cdot 2 x}

ⅰ

10 \cdot 2

20

\frac{\sqrt{5} \cdot x}{20 x}

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

x

\frac{\sqrt{5} \cdot x}{20 x}

\frac{\sqrt{5}}{20}

\frac{\sqrt{5}}{20}

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

The limit of

\sqrt{5}

\sqrt{5}

when

x \to 0

The limit of

\frac{\sqrt{5}}{20}

\frac{\sqrt{5}}{20}

when

x \to 0

So the limit of

\frac{\sqrt{5 + x} - \sqrt{5}}{2 x}

\frac{\sqrt{5}}{20}

when

x \to 0

▶Limit:

\frac{\sqrt{5}}{20}

4) The numeric value of:

\frac{\sqrt{5}}{20}

is:

\frac{\approx 2.2361}{20}

▷

\approx 0.1118