Chapter 14: Q. 38 (page 883)

In Problems 7– 42, find each limit algebraically.
$\lim_{x \to - 1} \frac{x^{3} + x^{2} + 3 x + 3}{x^{4} + x^{3} + 2 x + 2}$ .

Short Answer

Expert verified

The answer is 4.

Step by step solution

Step. 1 Given Information

Firstly, we check whether the given function is in indeterminant form or not.

$f (x) = \frac{x^{3} + x^{2} + 3 x + 3}{x^{4} + x^{3} + 2 x + 2}$

Put $x = - 1$ in the numerator we get,

$x^{3} + x^{2} + 3 x + 3 = {(- 1)}^{3} + {(- 1)}^{2} + 3 (- 1) + 3 = - 1 + 1 - 3 + 3 = 0 .$

Put $x = - 1$ in the denominator we get,

$x^{4} + x^{3} + 2 x + 2 = {(- 1)}^{4} + {(- 1)}^{3} + 2 (- 1) + 2 = 1 - 1 - 2 + 2 = 0 .$

Since both numerator and denominator gives 0 means they both have $x = - 1$ as their common root.

Step. 2 Factorizing

Denominator, $x^{4} + x^{3} + 2 x + 2 = x^{3} (x + 1) + 2 (x + 1) = (x^{3} + 2) (x + 1)$ .

Numerator , $x^{3} + x^{2} + 3 x + 3 = x^{2} (x + 1) + 3 (x + 1) = (x^{2} + 3) (x + 1) .$

So,

$\lim_{x \to - 1} \frac{x^{3} + x^{2} + 3 x + 3}{x^{4} + x^{3} + 2 x + 2} = \lim_{x \to - 1} \frac{(x^{2} + 3) (x + 1)}{(x^{3} + 2) (x + 1)} = \lim_{x \to - 1} \frac{x^{2} + 3}{x^{3} + 2} .$

Now we can put the limit directly.

Step. 3 Final calculation of the limit

$\lim_{x \to - 1} \frac{x^{2} + 3}{x^{3} + 2} = \frac{1 + 3}{- 1 + 2} = \frac{4}{1} = 4 .$

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In Problems 7– 42, find each limit algebraically.
$\lim_{x \to - 1} \frac{x^{3} + x^{2} + 3 x + 3}{x^{4} + x^{3} + 2 x + 2}$ .

Short Answer

Step by step solution

Step. 1 Given Information

Step. 2 Factorizing

Step. 3 Final calculation of the limit

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In Problems 7– 42, find each limit algebraically.limx→-1x3+x2+3x+3x4+x3+2x+2.

Short Answer

Step by step solution

Step. 1 Given Information

Step. 2 Factorizing

Step. 3 Final calculation of the limit

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Discrete Mathematics

Pure Maths

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Logic and Functions

Study anywhere. Anytime. Across all devices.

In Problems 7– 42, find each limit algebraically.
$\lim_{x \to - 1} \frac{x^{3} + x^{2} + 3 x + 3}{x^{4} + x^{3} + 2 x + 2}$ .