Evaluate parts A,B and C . EVALUATE (a)lem x² IXラー ×43x43メ+1(C) lem x2i×→ー x4+|

Name: Evaluate parts A,B and C . EVALUATE (a)lem x² IXラー ×43x43メ+1(C) lem x2
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Description: Evaluate parts A,B and C . EVALUATE (a)lem x² IXラー ×43x43メ+1(C) lem x2i×→ー x4+|

Given: (a) limx→-1- x2-1x3+3x2+3x+1(b) limx→-1+ x2-1x3+3x2+3x+1(c) limx→-1 x2-1x3+3x2+3x+1 To find:…

Answered: ' EVALUATE (a) lem x² I メ→ 43×キ3x+1」…

Calculus: Early Transcendentals

8th Edition

ISBN:9781285741550

Author:James Stewart

Publisher:James Stewart

Chapter1: Functions And Models

Section: Chapter Questions

Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...

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Evaluate parts A,B and C

. EVALUATE (a)
lem x² I
Xラー ×43x43メ+1
(C) lem x2i
×→ー x4+|

Expert Solution

Step 1

Given:

$(a) \lim_{x \to - 1^{-}} \frac{x^{2} - 1}{x^{3} + 3 x^{2} + 3 x + 1} (b) \lim_{x \to - 1^{+}} \frac{x^{2} - 1}{x^{3} + 3 x^{2} + 3 x + 1} (c) \lim_{x \to - 1} \frac{x^{2} - 1}{x^{3} + 3 x^{2} + 3 x + 1}$

To find:

Value of the limit.

Step 2

(a) $\lim_{x \to - 1^{-}} \frac{x^{2} - 1}{x^{3} + 3 x^{2} + 3 x + 1}$

Here as $x \to - 1^{-}$ we get the indeterminate form, so we can use the $L' H o s p i t a l' s$ rule to find the limit:

$\lim_{x \to a} \frac{f (x)}{g (x)} = \lim_{a \to a} \frac{f' (x)}{g' (x)}$

$\Rightarrow \lim_{x \to - 1^{-}} \frac{x^{2} - 1}{x^{3} + 3 x^{2} + 3 x + 1} = \lim_{x \to - 1^{-}} \frac{2 x}{3 x^{2} + 6 x + 3}$

$= 2 \cdot \lim_{x \to - 1^{-}} \frac{2 x}{3 x^{2} + 6 x + 3} = 2 \cdot \lim_{x \to - 1^{-}} \frac{x}{x (3 x + 6 + \frac{3}{x})} = 2 \cdot \lim_{x \to - 1^{-}} \frac{1}{(3 x + 6 + \frac{3}{x})}$

For x approaching -1 from left, $x < - 1$ $\Rightarrow 3 x + 6 + \frac{3}{x} < 0$

The denominator is a negative quantity approaching 0 from the left.

Therefore,

$\lim_{x \to - 1^{-}} \frac{x^{2} - 1}{x^{3} + 3 x^{2} + 3 x + 1} = 2 \cdot (- \infty) = - \infty$

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