Chapter 4: Q. 13 (page 193)

If $f (x) = - 2 x^{5} + x^{3} - 5 x^{2} + 7$ , then $\lim_{x \to - \infty} f (x) =________$ and $\lim_{x \to \infty} f (x) =________$ .

Short Answer

Expert verified

The solution is $\lim_{x \to - \infty} f (x) = \infty$ and $\lim_{x \to \infty} f (x) = - \infty$ .

Step by step solution

Given information

Function: $f (x) = - 2 x^{5} + x^{3} - 5 x^{2} + 7$

Calculation

We have,

$f (x) = - 2 x^{5} + x^{3} - 5 x^{2} + 7$

Now, $\lim_{x \to - \infty} f (x) = \lim_{x \to - \infty} (- 2 x^{5} + x^{3} - 5 x^{2} + 7)$

role="math" localid="1646902515040" $= \lim_{x \to - \infty} (x^{5}) (- 2 + \frac{1}{x^{2}} - \frac{5}{x^{3}} + \frac{7}{x^{5}})$

role="math" localid="1646902576542" $= \lim_{x \to - \infty} (x^{5}) (- 2 + 0 - 0 + 0)$

$= \infty$

and $\lim_{x \to \infty} f (x) = \lim_{x \to \infty} (- 2 x^{5} + x^{3} - 5 x^{2} + 7)$

$= \lim_{x \to \infty} (x^{5}) (- 2 + \frac{1}{x^{2}} - \frac{5}{x^{3}} + \frac{7}{x^{5}})$

$= \lim_{x \to \infty} (x^{5}) (- 2 + 0 - 0 + 0)$

$= - \infty$

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If $f (x) = - 2 x^{5} + x^{3} - 5 x^{2} + 7$ , then $\lim_{x \to - \infty} f (x) =________$ and $\lim_{x \to \infty} f (x) =________$ .

Short Answer

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Given information

Calculation

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If fx=-2x5+x3-5x2+7, then limx→-∞fx=________andlimx→∞fx=________.

Short Answer

Step by step solution

Given information

Calculation

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Discrete Mathematics

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Geometry

Applied Mathematics

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Decision Maths

Study anywhere. Anytime. Across all devices.

If $f (x) = - 2 x^{5} + x^{3} - 5 x^{2} + 7$ , then $\lim_{x \to - \infty} f (x) =________$ and $\lim_{x \to \infty} f (x) =________$ .