question 2 (1 point)\n$$\\lim_{x \\to \\pi} \\frac{-10 \\sin(x)}{x - \\pi}=$$ (enter a number for your…

question 2 (1 point)\n$$\\lim_{x \\to \\pi} \\frac{-10 \\sin(x)}{x - \\pi}=$$ (enter a number for your answer.)\n> view hint for question 2\nquestion 3 (1 point)\nuse lhospitals rule to determine which of the following equal 0 (select all that apply):\n$$\\lim_{x \\to \\infty} \\frac{x^{2}}{\\ln(x)}$$\n$$\\lim_{x \\to \\infty} \\frac{5 e^{x}}{13 \\sqrt{x}}$$\n$$\\lim_{x \\to \\infty} \\frac{\\ln(x^{10})}{3 x}$$\n$$\\lim_{x \\to \\infty} \\frac{10 x^{4}}{e^{x}}$$

question 2 (1 point)\n$$\\lim_{x \\to \\pi} \\frac{-10 \\sin(x)}{x - \\pi}=$$ (enter a number for your answer.)\n> view hint for question 2\nquestion 3 (1 point)\nuse lhospitals rule to determine which of the following equal 0 (select all that apply):\n$$\\lim_{x \\to \\infty} \\frac{x^{2}}{\\ln(x)}$$\n$$\\lim_{x \\to \\infty} \\frac{5 e^{x}}{13 \\sqrt{x}}$$\n$$\\lim_{x \\to \\infty} \\frac{\\ln(x^{10})}{3 x}$$\n$$\\lim_{x \\to \\infty} \\frac{10 x^{4}}{e^{x}}$$

Answer

Explanation:

Step1: Use L'Hospital's Rule

When (x\to\pi), (\frac{- 10\sin(x)}{x - \pi}) is in the (\frac{0}{0}) form. By L'Hospital's Rule, (\lim_{x\to\pi}\frac{-10\sin(x)}{x - \pi}=\lim_{x\to\pi}\frac{-10\cos(x)}{1}).

Step2: Substitute (x = \pi)

Substitute (x=\pi) into (\frac{-10\cos(x)}{1}), we get (\frac{-10\cos(\pi)}{1}). Since (\cos(\pi)=-1), then (\frac{-10\times(-1)}{1}=10).

Answer:

(10)

For Question 3:

For (\lim_{x\to\infty}\frac{x^{2}}{\ln(x)})

It is in the (\frac{\infty}{\infty}) form. By L'Hospital's Rule, (\lim_{x\to\infty}\frac{x^{2}}{\ln(x)}=\lim_{x\to\infty}\frac{2x}{\frac{1}{x}}=\lim_{x\to\infty}2x^{2}=\infty)

For (\lim_{x\to\infty}\frac{5e^{x}}{13\sqrt{x}})

It is in the (\frac{\infty}{\infty}) form. By L'Hospital's Rule, (\lim_{x\to\infty}\frac{5e^{x}}{13\sqrt{x}}=\lim_{x\to\infty}\frac{5e^{x}}{\frac{13}{2\sqrt{x}}}=\lim_{x\to\infty}\frac{10e^{x}\sqrt{x}}{13}=\infty)

For (\lim_{x\to\infty}\frac{\ln(x^{10})}{3x})

First, simplify (\ln(x^{10}) = 10\ln(x)). So (\lim_{x\to\infty}\frac{\ln(x^{10})}{3x}=\lim_{x\to\infty}\frac{10\ln(x)}{3x}). It is in the (\frac{\infty}{\infty}) form. By L'Hospital's Rule, (\lim_{x\to\infty}\frac{10\ln(x)}{3x}=\lim_{x\to\infty}\frac{\frac{10}{x}}{3}=\lim_{x\to\infty}\frac{10}{3x}=0)

For (\lim_{x\to\infty}\frac{10x^{4}}{e^{x}})

It is in the (\frac{\infty}{\infty}) form. By L'Hospital's Rule, (\lim_{x\to\infty}\frac{10x^{4}}{e^{x}}=\lim_{x\to\infty}\frac{40x^{3}}{e^{x}}). Still (\frac{\infty}{\infty}) form. Apply L'Hospital's Rule again: (\lim_{x\to\infty}\frac{120x^{2}}{e^{x}}). Again (\frac{\infty}{\infty}) form. Apply L'Hospital's Rule: (\lim_{x\to\infty}\frac{240x}{e^{x}}). Again (\frac{\infty}{\infty}) form. Apply L'Hospital's Rule: (\lim_{x\to\infty}\frac{240}{e^{x}} = 0)

Answer:

(\lim_{x\to\infty}\frac{\ln(x^{10})}{3x}=0), (\lim_{x\to\infty}\frac{10x^{4}}{e^{x}}=0)