Chapter 2.3, Problem 29E

### Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343

Chapter
Section

### Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343
Textbook Problem

# Evaluate the limit, if it exists. lim t → 0 ( 1 t 1 + t − 1 t )

To determine

To evaluate: The limit of the function limt0(1t1+t1t).

Explanation

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 square formula: (a2b2)=(a+b)(ab)

Fact 1:

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

Let f(t)=1t1+t1t (1)

Note 1:

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

f(0)=1(0)1+010=110=00

Note 2:

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

limt0(t2(1+t))=limt0t2limt0(1+t) (by limit law 4)=limt0t2(limt0(1+t)) (by limit law 11)=limt0t2(limt01+limt0t) (by limit law 1)=02(1+0) (by limit law 9,7,8)

=0

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 t approaches 0 but t0.

Simplify f(t) by using elementary algebra.

f(t)=1t1+t1t

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

f(t)=1t1+t1t=tt1+tt21+t=t(11+t)t21+t

Take the conjugate of the numerator and multiply and divide by f(t)

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