Answered: if x+ 1 f(x) = { x* a = 1 1 if x= 1

Algebra & Trigonometry with Analytic Geometry

13th Edition

ISBN:9781133382119

Author:Swokowski

Publisher:Swokowski

Chapter5: Inverse, Exponential, And Logarithmic Functions

Section5.3: The Natural Exponential Function

Problem 54E

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Question

Explain why the function is discontinuous at the given
number a . Sketch the graph of the function.

Expert Solution

Step 1

The function is given ,

$f (x) = (\begin{matrix} \frac{x^{2} - x}{x^{2} - 1}, i f x \neq 1 \\ 1, i f x = 1 \end{matrix})$ for $a = 1 .$

To check the function discontinuity,

The left hand limit is,

$\begin{array}{rcl} \lim_{h \to 0} f (1 - h) & = & \lim_{h \to 0} \frac{{(1 - h)}^{2} - (1 - h)}{{(1 - h)}^{2} - 1} \\ = & \lim_{h \to 0} [\frac{1 + h^{2} - 2 h - 1 + h}{1 + h^{2} - 2 h - 1}] \\ = & \lim_{h \to 0} [\frac{h^{2} - h}{h^{2} - 2 h}] \\ = & \lim_{h \to 0} [\frac{h - 1}{h - 2}] \\ = & \frac{1}{2} \end{array}$

Step 2

The right hand limit is,

$\begin{array}{rcl} \lim_{h \to 0} f (1 + h) & = & \lim_{h \to 0} \frac{{(1 + h)}^{2} - (1 + h)}{{(1 + h)}^{2} - 1} \\ = & \lim_{h \to 0} [\frac{1 + h^{2} + 2 h - 1 - h}{1 + h^{2} + 2 h - 1}] \\ = & \lim_{h \to 0} [\frac{h^{2} + h}{h^{2} + 2 h}] \\ = & \lim_{h \to 0} [\frac{h + 1}{h + 2}] \\ = & \frac{1}{2} \end{array}$

The function at $a = 1$ ,is $f (1) = 1 .$

Thus , the LHL $=$ RHL $≅ f (1)$ .

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