   Chapter 2.3, Problem 32E

Chapter
Section
Textbook Problem

Evaluate the limit, if it exists. lim h → 0 1 ( x − h ) 2 − 1 x 2 h

To determine

To evaluate: The limit of the function limh01(x+h)21x2h.

Explanation

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.

Let f(h)=1(x+h)21x2h (1)

Note 1:

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

f(0)=1(x+0)21x20=1x21x20=00

Note 2:

The Quotient rule is not applicable for the function f(h) as the limit of the denominator is zero.

limh0(h)=0 (by limit law 8)

Note 3:

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

Calculation:

By note 3, take the limit h approaches to 0 but h0.

Simplify f(h) by using elementary algebra.

f(h)=1(x+h)21x2h

Simplify f(h) by the method of cross multiplication,

f(h)=x2(x+h)2x2(x+h)2h=<

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