/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 24 Determine the following limits. ... [FREE SOLUTION] | 91影视

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Chapter 2: Problem 24

Determine the following limits. $$\lim _{x \rightarrow-\infty}\left(2 x^{-8}+4 x^{3}\right)$$

Short Answer

Expert verified
Answer: The limit of the function as \(x\) approaches negative infinity is \(-\infty\).

Step by step solution

01

Find the limit of the first term

The first term of the function is \(2x^{-8}\). To find the limit as x approaches negative infinity, we need to determine how this term behaves when x decreases: $$\lim_{x \rightarrow -\infty} 2x^{-8} = \lim_{x \rightarrow -\infty} \frac{2}{x^8}$$ As x approaches negative infinity, the denominator \(x^8\) will approach infinity, so the fraction becomes smaller and smaller. Consequently: $$\lim_{x \rightarrow -\infty} \frac{2}{x^8} = 0$$
02

Find the limit of the second term

The second term of the function is \(4x^3\). To find the limit as x approaches negative infinity, we need to determine how this term behaves when x decreases: $$\lim_{x \rightarrow -\infty} 4x^3$$ As x approaches negative infinity, \(x^3\) will also approach negative infinity, so the entire term will approach negative infinity: $$\lim_{x \rightarrow -\infty} 4x^3 = -\infty$$
03

Find the overall limit of the function

Now that we have the limits of both terms, we can find the overall limit of the function: $$\lim _{x \rightarrow-\infty}\left(2 x^{-8}+4 x^{3}\right) = \lim_{x \rightarrow -\infty} 2x^{-8} + \lim_{x \rightarrow -\infty} 4x^3$$ Since we found that the limit of the first term is 0 and the limit of the second term is \(-\infty\), the overall limit of the function is: $$\lim _{x \rightarrow-\infty}\left(2 x^{-8}+4 x^{3}\right) = 0 + (-\infty) = -\infty$$

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Most popular questions from this chapter

Let \(f(x)=\frac{|x|}{x},\) for \(x \neq 0\) a. Sketch a graph of \(f\) on the interval [-2,2] b. Does \(\lim _{x \rightarrow 0} f(x)\) exist? Explain your reasoning after first examining \(\lim _{x \rightarrow 0^{-}} f(x)\) and \(\lim _{x \rightarrow 0^{+}} f(x)\)

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Sketching graphs Sketch a possible graph of a function \(f\) that satisfies all the given conditions. Be sure to identify all vertical and horizontal asymptotes. $$\begin{aligned} &f(-1)=-2, f(1)=2, f(0)=0, \lim _{x \rightarrow \infty} f(x)=1\\\ &\lim _{x \rightarrow-\infty} f(x)=-1 \end{aligned}$$

Use the following definitions. Assume fexists for all \(x\) near a with \(x>\) a. We say the limit of \(f(x)\) as \(x\) approaches a from the right of a is \(L\) and write \(\lim _{x \rightarrow a^{+}} f(x)=L,\) if for any \(\varepsilon>0\) there exists \(\delta>0\) such that $$|f(x)-L|<\varepsilon \quad \text { whenever } \quad 0< x-a<\delta$$ Assume fexists for all \(x\) near a with \(x < \) a. We say the limit of \(f(x)\) as \(x\) approaches a from the left of a is \(L\) and write \(\lim _{x \rightarrow a^{-}} f(x)=L,\) if for any \(\varepsilon > 0 \) there exists \(\delta > 0\) such that $$|f(x)-L| < \varepsilon \quad \text { whenever } \quad 0< a-x <\delta$$ Prove the following limits for $$f(x)=\left\\{\begin{array}{ll} 3 x-4 & \text { if } x<0 \\ 2 x-4 & \text { if } x \geq 0 \end{array}\right.$$ a. \(\lim _{x \rightarrow 0^{+}} f(x)=-4\) b. \(\lim _{x \rightarrow 0^{-}} f(x)=-4\) c. \(\lim _{x \rightarrow 0} f(x)=-4\)

Use the following definitions. Assume fexists for all \(x\) near a with \(x>\) a. We say the limit of \(f(x)\) as \(x\) approaches a from the right of a is \(L\) and write \(\lim _{x \rightarrow a^{+}} f(x)=L,\) if for any \(\varepsilon>0\) there exists \(\delta>0\) such that $$|f(x)-L|<\varepsilon \quad \text { whenever } \quad 0< x-a<\delta$$ Assume fexists for all \(x\) near a with \(x < \) a. We say the limit of \(f(x)\) as \(x\) approaches a from the left of a is \(L\) and write \(\lim _{x \rightarrow a^{-}} f(x)=L,\) if for any \(\varepsilon > 0 \) there exists \(\delta > 0\) such that $$|f(x)-L| < \varepsilon \quad \text { whenever } \quad 0< a-x <\delta$$ Prove that \(\lim _{x \rightarrow 0^{+}} \sqrt{x}=0\).
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