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Precalculus Enhanced with Graphing Utilities

Sullivan

$Math Studyset 91影视 Explanations$ Math

6th Edition 路 1200 pages

ISBN: 9780321795465

Chapter 14: Q. 11 (page 905)

find the limit $\lim_{x 鈫� 3} \frac{x^{4} - 3 x^{3} + x - 3}{x^{3} - 3 x^{2} + 2 x - 6}$

Short Answer

Expert verified

$\lim_{x 鈫� 3} \frac{x^{4} - 3 x^{3} + x - 3}{x^{3} - 3 x^{2} + 2 x - 6} = \frac{28}{11}$

Step by step solution

01

Step 1. Given information

$\lim_{x 鈫� 3} \frac{x^{4} - 3 x^{3} + x - 3}{x^{3} - 3 x^{2} + 2 x - 6}$

02

Step 2. limit of the function

We can find limit of rational function by $\lim_{x 鈫� c} f (x) = f (c)$ if c is in the domain of f(x) and in the domain of f(x)= $\frac{x^{4} - 3 x^{3} + x - 3}{x^{3} - 3 x^{2} + 2 x - 6}$ 3 is not included So we have to simplify the f(x) for finding the limit at 3.

$\lim_{x 鈫� 3} \frac{x^{4} - 3 x^{3} + x - 3}{x^{3} - 3 x^{2} + 2 x - 6} = \lim_{x 鈫� 3} \frac{(x^{3} + 1) (x - 3)}{(x^{2} + 2) (x - 3)} = \lim_{x 鈫� 3} \frac{x^{3} + 1}{x^{2} + 2} = \frac{(3)^{3} + 1}{(3)^{2} + 2} = \frac{28}{11}$

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