# The value of lim u → 1 u 4 − 1 u 3 + 5 u 2 − 6 u .

### 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 11RE
To determine

## To find: The value of limu→1u4−1u3+5u2−6u.

Expert Solution

The limit of the function is 47_.

### Explanation of Solution

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).

Difference of squared formula: (a2b2)=(a+b)(ab)

Fact 1:

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

Evaluation:

Let f(u)=u41u3+5u26u (1)

Note 1:

The direct substitution method is not applicable for the function f(u) as the function f(1) is in an indeterminate form when u=1.

f(1)=(1)41(1)3+5(1)26(1)=111+5(1)6=066=00

Note 2:

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

Calculation:

By note 2, consider the limit u approaches 1 but u1.

Simplify f(u) by using elementary algebra as follows.

f(u)=u41u3+5u26u=(u2)2(1)2u3+5u26u

Apply the difference of squared formula,

f(u)=(u2+1)(u21)(u3+5u26u)

Again, apply the difference of squared formula in the numerator,

f(u)=(u2+1)(u+1)(u1)(u3+5u26u) (2)

Factorize the denominator of f(x),

u3+5u26u=u(u2+5u6)=u(u2+6uu6)=u(u(u+6)(u+6))=u(u+6)(u1)

Substitute u(u+6)(u1) for u3+5u26u in equation (2),

f(u)=(u2+1)(u+1)(u1)u(u+6)(u1)

Since the limit u approaches 1 but not equal to 1, cancel the common term u10 from both the numerator and the denominator,

f(u)=(u2+1)(u+1)u(u+6)

Use fact 1, f(u)=(u2+1)(u+1)u(u+6) and u1, then limu1u41u3+5u26u=limu1(u2+1)(u+1)u(u+6).

Apply the direct substitution property on the limit function.

limu1(u2+1)(u+1)u(u+6)=((1)2+1)((1)+1)(1)((1)+6)=(1+1)(1+1)(1)(1+6)=(2)(2)(1)(7)=47

Thus, the limit of the function is 47_.

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