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Chapter 1: Q. 6 (page 153)

Calculating limits: Find each limit by hand.
$\lim_{x 鈫� - 鈭�} \frac{x^{3} + 2 x + 1}{1 - x^{4}} .$

Short Answer

Expert verified

The answer of the limit is 0.

Step by step solution

01

Step 1. Given Information.

Given: $\lim_{x 鈫� - 鈭�} \frac{x^{3} + 2 x + 1}{1 - x^{4}}$ .

02

Step 2. Calculating the limit using direct substitution.

$S o l v i n g b y h a n d, \lim_{x 鈫� - 鈭�} \frac{x^{3} + 2 x + 1}{1 - x^{4}} D i v i d i n g b o t h n u m e r a t o r a n d d e n o m i n a t o r b y x^{4} . = \lim_{x 鈫� - 鈭�} \frac{\frac{1}{x} + \frac{2}{x^{3}} + \frac{1}{x^{4}}}{\frac{1}{x^{4}} - 1} = \frac{\frac{1}{鈭�} + \frac{2}{鈭�} + \frac{1}{鈭�}}{\frac{1}{鈭�} - 1} = \frac{0 + 0 + 0}{0 - 1} = 0 .$

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Most popular questions from this chapter

Write delta-epsilon proofs for each of the limit statements $\lim_{x 鈫� c} f (x) = L$ in Exercises $47 - 60$ .

$\lim_{x 鈫� - 6} (x + 2) = - 4$ .

$\lim_{x 鈫� 蟺} \frac{1}{\cos e c (蟺 - x)}$

Calculate each of the limits:

$\lim_{h 鈫� 0} \frac{{(3 + h)}^{2} - 3^{2}}{h}$ .

For each limit $\underset{x 鈫� c}{l i m} f (x) = L$ in Exercises 43鈥�54, use graphs and algebra to approximate the largest value of $未$ such that if $x 鈭� (c - 未, c) 鈭� (c, c + 未) then f (x) 鈭� (L - 蔚, L + 蔚)$

$\underset{x 鈫� - 1}{l i m} \frac{x^{2} - 2 x - 3}{x + 1} = - 4, 蔚 = 0.1$

Write a delta鈥揺psilon proof that proves that f is continuous on its domain. In each case, you will need to assume that 未 is less than or equal to 1.

$f (x) = x^{鈭� 1}$

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