# The limit of the function lim h → 0 ( h − 1 ) 3 + 1 h .

### Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805

### Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805

#### Solutions

Chapter 2, Problem 7RE
To determine

## To evaluate: The limit of the function limh→0(h−1)3+1h.

Expert Solution

The limit of the function is 3.

### Explanation of Solution

Formula used:

Limit Laws:

Suppose that c is a constant and the limits limxaf(x) and limxag(x) exist, then

Limit law 8: limxax=a

Direct substitution property:

If f is a polynomial or a rational function and a is in the domain of f, then limxaf(x)=f(a).

Fact 1:

If f(x)=g(x) when xa, then limxaf(x)=limxag(x), provided the limit exist.

Given:

Let f(h)=(h1)3+1h (1)

Note 1:

The direct substitution method is not applicable for the function f(h) as the function f(0) is in an indeterminate form when h=0.

f(0)=(01)3+10=1+10=00

Note 2:

The limit may be infinite or some finite value when both the numerator and the denominator approach to 0.”

Calculation:

By note 2, consider the limit h approaches to zero but h0.

Simplify f(h) by using elementary algebra.

f(h)=(h1)3+1h

Expand the numerator and simplify further,

f(h)=(h33h2+3h1)+1h=h33h2+3h1+1h=h33h2+3hh=h(h23h+3)h

Since the limit h approaches zero but not equal to zero, cancel the common term h from both the numerator and the denominator,

f(h)=(h23h+3)

By fact 1, if f(h)=h23h+3 and h0, then limh0(h1)3+1h=limh0(h23h+3).

Apply the direct substitution property on the limit function.

limh0(h23h+3)=(023(0)+3)=3

Thus, the limit of the function is 3_.

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