# Finding a Limit In Exercises 39–-50, find the limit (if it exists). If it does not exist, explain why. lim x → 3 + 1 x + 3 lim x → 6 − x − 6 x 2 − 36 lim x → 25 + x − 5 x − 25 lim x → 3 − | x − 3 | x − 3 lim x → 2 f ( x ) , where f ( x ) = { ( x − 2 ) 2 , x ≤ 2 2 − x , x &gt; 2 lim x → 1 + g ( x ) , where g ( x ) = { 1 − x , x ≤ 1 x + 1 , x &gt; 1 lim t → 1 h ( t ) , where h ( t ) = { t 3 + 1 , t &lt; 1 1 2 ( t + 1 ) , t ≥ 1 lim x → − 2 f ( s ) , where f ( s ) = { − s 2 − 4 s − 2 , s ≤ − 2 s 2 + 4 s + 6 , s &gt; − 2 lim x → 2 − ( 2 [ x ] + 1 ) lim x → 4 [ [ x − 1 ] ] lim x → 2 − x 2 − 4 | x − 2 | lim x → 1 + x ( x − 1 )

### Calculus: Early Transcendental Fun...

7th Edition
Ron Larson + 1 other
Publisher: Cengage Learning
ISBN: 9781337552516

Chapter
Section

### Calculus: Early Transcendental Fun...

7th Edition
Ron Larson + 1 other
Publisher: Cengage Learning
ISBN: 9781337552516
Chapter 2, Problem 42RE
Textbook Problem
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## Finding a Limit In Exercises 39–-50, find the limit (if it exists). If it does not exist, explain why. lim x → 3 + 1 x + 3 lim x → 6 − x − 6 x 2 − 36 lim x → 25 + x − 5 x − 25 lim x → 3 − | x − 3 | x − 3 lim x → 2 f ( x ) , where f ( x ) = { ( x − 2 ) 2 , x ≤ 2 2 − x , x > 2 lim x → 1 + g ( x ) , where g ( x ) = { 1 − x , x ≤ 1 x + 1 , x > 1 lim t → 1 h ( t ) , where h ( t ) = { t 3 + 1 , t < 1 1 2 ( t + 1 ) , t ≥ 1 lim x → − 2 f ( s ) , where f ( s ) = { − s 2 − 4 s − 2 , s ≤ − 2 s 2 + 4 s + 6 , s > − 2 lim x → 2 − ( 2 [ x ] + 1 ) lim x → 4 [ [ x − 1 ] ] lim x → 2 − x 2 − 4 | x − 2 | lim x → 1 + x ( x − 1 )

To determine
The value of the limit (if exists), limx3|x3|x3, where || denotes the modulus function.

### Explanation of Solution

Given:

The expression, limx3|x3|x3.

Explanation:

Consider the expression, limx3|x3|x3.

First, draw the graph for the function f(x)=|x3|x3

The domain of f(x) is all real numbers except x=3.

For x>3,

|x3|=x3

So,

f(x)=x3x3=

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