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Chapter 1: Problem 33

Find the limit (if it exists). $$ \lim _{x \rightarrow 0} \frac{[1 /(3+x)]-(1 / 3)}{x} $$

Short Answer

Expert verified
The limit of the function, \(\lim_{x \rightarrow 0}\frac{[1/(3+x)]-(1/3)}{x} = \frac{1}{9}\.

Step by step solution

01

Combine fractions under one denominator

First, combine the fractions under one denominator. To do this, find a common denominator, which is \(3(3 + x)\) and rewrite the fractions as \(\frac{3 - (3 + x)}{3*(3 + x)} = \frac{-x}{3*(3 + x)}\). So, the limit now is \(\lim_{x \rightarrow 0}\frac{-x}{3*(3 + x)*x}\).
02

Simplify equation

Simplify the equation by cancelling -x in both the numerator and denominator. Now we get \(\lim_{x \rightarrow 0}\frac{1}{3*(3 + x)}\).
03

Substitute x=0 in the function.

Substitute \(x=0\) in the function \(\frac{1}{3*(3 + x)}\). It results in \(\frac{1}{3*3}= \frac{1}{9}\).

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