Q. 49 Calculate each limit in Exercise... [FREE SOLUTION]

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Chapter 1: Q. 49 (page 149)

Calculate each limit in Exercises 35鈥�80.
$\lim_{x 鈫� 鈭�} \frac{{(3 x + 1)}^{2} (x - 1)}{1 - x^{3}}$

Short Answer

Expert verified

The limit is - 9

Step by step solution

Step 1. Given Information

Given expression: $\lim_{x 鈫� 鈭�} \frac{{(3 x + 1)}^{2} (x - 1)}{1 - x^{3}}$

We want to find limit of given expression.

Step 2. Solution:

Simplify:

$\lim_{x 鈫� 鈭�} \frac{{(3 x + 1)}^{2} (x - 1)}{1 - x^{3}}$

We know that $a^{3} - b^{3} = (a - b) (a^{2} + b^{2} + a b) a n d {(a + b)}^{2} = a^{2} + 2 a b + b^{2}$

So we have

$\lim_{x 鈫� 鈭�} \frac{(9 x^{2} + 6 x + 1) (x - 1)}{(1 - x) (1 + x^{2} + x)} = \lim_{x 鈫� 鈭�} - \frac{(9 x^{2} + 6 x + 1)}{(1 + x^{2} + x)}$

Take x² Common from numerator and denominator we get

$\lim_{x 鈫� 鈭�} - \frac{x^{2} (9 + \frac{6}{x} + \frac{1}{x^{2}})}{x^{2} (\frac{1}{x^{2}} + 1 + \frac{1}{x})} = \lim_{x 鈫� 鈭�} - \frac{9 + \frac{6}{x} + \frac{1}{x^{2}}}{\frac{1}{x^{2}} + 1 + \frac{1}{x}}$

Now take limit we get:

$\lim_{x 鈫� 鈭�} - \frac{9 + \frac{6}{x} + \frac{1}{x^{2}}}{\frac{1}{x^{2}} + 1 + \frac{1}{x}} = - \frac{9 + \frac{6}{鈭�} + \frac{1}{{鈭�}^{2}}}{\frac{1}{{鈭�}^{2}} + 1 + \frac{1}{鈭�}} = - \frac{9}{1} = - 9$

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Calculate each limit in Exercises 35鈥�80.
$\lim_{x 鈫� 鈭�} \frac{{(3 x + 1)}^{2} (x - 1)}{1 - x^{3}}$

Short Answer

Step by step solution

Step 1. Given Information

Step 2. Solution:

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91影视

Calculate each limit in Exercises 35鈥�80. limx鈫�鈭�(3x+1)2(x-1)1-x3

Short Answer

Step by step solution

Step 1. Given Information

Step 2. Solution:

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Most popular questions from this chapter

Recommended explanations on Math Textbooks

Probability and Statistics

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Theoretical and Mathematical Physics

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Study anywhere. Anytime. Across all devices.

Calculate each limit in Exercises 35鈥�80.
$\lim_{x 鈫� 鈭�} \frac{{(3 x + 1)}^{2} (x - 1)}{1 - x^{3}}$