limit as x approaches 1- of (x^3-1)/(x-1)

1) Compute the limit of

\frac{x^{3} - 1}{x - 1}

x \to 1

⁻

ⓘ The power with a constant exponent is continuous here: substitute the limit of the base

The limit of

x^{3}

1

when

x \to 1

⁻

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

The limit of

x^{3} - 1

is (

1 - 1

)⁻ when

x \to 1

⁻

⇥1.1) Let's calculate:

1 - 1

⇥ⓘ Start from

1

and move back

1

on the number line, down to zero:

1 - 1

0

⇥

0

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

The limit of

x - 1

is (

1 - 1

)⁻ when

x \to 1

⁻

⇥1.2) Let's evaluate:

1 - 1

⇥ⓘ Start from

1

and move back

1

on the number line, down to zero:

1 - 1

0

⇥

0

For

\frac{x^{3} - 1}{x - 1}

, we encounter the indeterminate form 0/0 as

x

tends to

1

2) Let's apply L'Hôpital's rule (limit of a/b = limit of a'/b') by differentiating numerator and denominator:

3) Let's perform the differentiation:

\frac{d}{d x} (x^{3} - 1)

ⓘ The derivative of a sum is the sum of the derivatives:

\frac{d}{d x} (x^{3} - 1)

\frac{d}{d x} (x^{3}) + 0

\frac{d}{d x} (x^{3}) + 0

ⓘ Adding 0 changes nothing:

\frac{d}{d x} (x^{3}) + 0

\frac{d}{d x} (x^{3})

■The derivative of x^n is n*x^(n-1):

\frac{d}{d x} (x^{3})

3 x^{2}

3 x^{2}

4) Let's differentiate:

\frac{d}{d x} (x - 1)

ⓘ The derivative of a sum is the sum of the derivatives:

\frac{d}{d x} (x - 1)

1 + 0

1 + 0

ⓘ Adding 0 changes nothing:

1 + 0

1

1

Using those derivatives, we obtain a new fraction whose limit can be computed:

ⓘ The power with a constant exponent is continuous here: substitute the limit of the base

The limit of

x^{2}

1

when

x \to 1

⁻

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

The limit of

3 x^{2}

3

when

x \to 1

⁻

So the limit of

\frac{x^{3} - 1}{x - 1}

3

when

x \to 1

⁻

▶Limit:

3