   Chapter 2.5, Problem 37E ### Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343

#### Solutions

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

8th Edition
James Stewart
ISBN: 9781305270343
Textbook Problem

# Use continuity to evaluate the limit. lim x → 1 ln ( 5 − x 2 1 + x )

To determine

To evaluate: The limit of the function f(x)=ln(5x21+x) as x approaches 1 by using continuity and check the function is continuous or not.

Explanation

Definition used: “A function f is continuous at a number a if limxaf(x)=f(a)”.

Theorems used:

7. The functions such as “Polynomials, rational functions, root functions, trigonometric functions, inverse trigonometric functions, exponential functions and logarithmic functions” are continuous at every number in their domains.

9. If g is continuous at a and f is continuous at g(a), then the composite function fg given by (fg)(x)=f(g(x)) is continuous at a.

The domain is the set of all input values of the function for which the function is real and defined.

Calculation:

Consider the function f(x) is composition of two functions. That is, f(x)=h(g(x)) where h(x)=lnx and g(x)=5x21+x.

The rational function g(x)=5x21+x is defined for every real numbers except x1. Also the function f(x)=ln(5x21+x) is defined whenever 5x21+x must be greater than zero and 1+x is not equal to zero.

Thus, the function f(x) is defined on the interval (,5)(1,5).

Thus, the domain of the function f(x) is D=(,5)(1,5)

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