Question

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

view hint for question 2

question 3 (1 point)
use lhospitals rule to determine which of the following equal 0 (select all that apply):
$$\lim_{x \to \infty} \frac{x^{2}}{\ln(x)}$$
$$\lim_{x \to \infty} \frac{5 e^{x}}{13 \sqrt{x}}$$
$$\lim_{x \to \infty} \frac{\ln(x^{10})}{3 x}$$
$$\lim_{x \to \infty} \frac{10 x^{4}}{e^{x}}$$

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: \(\lim_{x\to\infty}\frac{240}{e^{x}} = 0\)