Question

question 5 1 pts \\( \lim_{x \to \infty} \frac{6x^3 + \arctan x}{1 - 3x^3} \\) \\( \circ \\) \\( \pi/2 \\) \\( \circ \\) -2 \\( \circ \\) dne \\( \circ \\) 2 question 6 1 pts \\( \lim_{x \to -\infty} \frac{2e^{2x} + e^{-x}}{e^{2x} + 3e^{-x}} \\) \\( \circ \\) 1/3 \\( \circ \\) 0 \\( \circ \\) 2 \\( \circ \\) dne

Explanation:

Question 5

Step1: Divide numerator and denominator by \(x^3\)

When \(x \to \infty\), the highest power of \(x\) in the numerator and denominator is \(x^3\). So we divide each term in the numerator and the denominator by \(x^3\):
\[
\lim_{x \to \infty} \frac{\frac{6x^3}{x^3} + \frac{\arctan x}{x^3}}{\frac{1}{x^3} - \frac{3x^3}{x^3}}=\lim_{x \to \infty} \frac{6 + \frac{\arctan x}{x^3}}{\frac{1}{x^3} - 3}
\]

Step2: Evaluate limits of each term

We know that \(\lim_{x \to \infty}\arctan x=\frac{\pi}{2}\), and \(\lim_{x \to \infty}\frac{\arctan x}{x^3} = 0\) (because the numerator is bounded and the denominator goes to \(\infty\)), and \(\lim_{x \to \infty}\frac{1}{x^3}=0\). Substituting these limits into the expression:
\[
\frac{\lim_{x \to \infty}6+\lim_{x \to \infty}\frac{\arctan x}{x^3}}{\lim_{x \to \infty}\frac{1}{x^3}-\lim_{x \to \infty}3}=\frac{6 + 0}{0 - 3}=- 2
\]

Step1: Analyze the behavior of exponential functions as \(x\to-\infty\)

Recall that for exponential functions, when \(x\to-\infty\), \(e^{ax}\) behaves differently depending on the sign of \(a\). If \(a>0\), then \(e^{ax}\to0\) as \(x\to-\infty\); if \(a < 0\), then \(e^{ax}\to\infty\) as \(x\to-\infty\).

For the given limit \(\lim_{x\to-\infty}\frac{2e^{2x}+e^{-x}}{e^{2x}+3e^{-x}}\), let's rewrite \(e^{-x}\) as \(e^{(-1)x}\) and \(e^{2x}\) as \(e^{(2)x}\). When \(x\to-\infty\), \(2x\to-\infty\) (since \(2>0\)) and \(-x\to\infty\) (since \(- 1<0\)). So \(e^{2x}\to0\) and \(e^{-x}\to\infty\) as \(x\to-\infty\).

We can divide both the numerator and the denominator by \(e^{-x}\) (the dominant term in the numerator and denominator as \(x\to-\infty\)):
\[
\lim_{x\to-\infty}\frac{\frac{2e^{2x}}{e^{-x}}+\frac{e^{-x}}{e^{-x}}}{\frac{e^{2x}}{e^{-x}}+\frac{3e^{-x}}{e^{-x}}}=\lim_{x\to-\infty}\frac{2e^{3x}+1}{e^{3x}+3}
\]

Step2: Evaluate the limit

Now, as \(x\to-\infty\), \(3x\to-\infty\), so \(e^{3x}\to0\). Substituting this limit into the expression:
\[
\frac{2\times0 + 1}{0+3}=\frac{1}{3}
\]

Answer:

\(-2\) (corresponding to the option " -2")

Question 6