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

1) Compute the limit of

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

t \to 1

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

The limit of

t^{3}

1

when

t \to 1

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

The limit of

t^{3} - 1

1 - 1

when

t \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

t - 1

1 - 1

when

t \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{t^{3} - 1}{t - 1}

, the limit takes the indeterminate form 0/0 when

t

→

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 differentiate:

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

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

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

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

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

ⓘ Adding 0 changes nothing:

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

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

■The derivative of x^n is n*x^(n-1):

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

3 t^{2}

3 t^{2}

4) Let's differentiate:

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

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

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

1 + 0

1 + 0

ⓘ Adding 0 changes nothing:

1 + 0

1

1

With the derivatives of the numerator and the denominator, we obtain a new fraction whose limit can be determined:

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

The limit of

t^{2}

1

when

t \to 1

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

The limit of

3 t^{2}

3

when

t \to 1

So the limit of

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

3

when

t \to 1

▶Limit:

3