## University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson

# Chapter 2 - Section 2.2 - Limit of a Function and Limit Laws - Exercises - Page 68: 70

#### Answer

$$\lim_{x\to3}h(x)=2$$

#### Work Step by Step

$$h(x)=\frac{x^2-2x-3}{x^2-4x+3}$$ (a) The table is shown below. As we can see from 2 tables below, as $x$ gets more decimals and approaches $3$, the value of $h(x)$ also gets more decimals and apparently approaches $2$. So I would estimate here that $\lim_{x\to3}h(x)=2$. (b) The graph is shown below. Again, looking at the graph, the closer $x$ approaches $3$, the closer $h(x)$ approaches $2$. (c) $$\lim_{x\to3}h(x)=\lim_{x\to3}\frac{x^2-2x-3}{x^2-4x+3}=\lim_{x\to3}\frac{(x+1)(x-3)}{(x-1)(x-3)}$$ $$\lim_{x\to3}h(x)=\lim_{x\to3}\frac{x+1}{x-1}=\frac{3+1}{3-1}=\frac{4}{2}=2$$ Therefore, $$\lim_{x\to3}h(x)=2$$

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