Math / Calculus

QuestionTrue or False $\quad$ NO 10 $x=0$ 为 $y=\sin \frac{1}{x}$ 的振荡间断点（） $\text { 2. } x=\pi / 2 \text { 为 } y=\tan x \text { 的无穷间断 }$

Studdy Solution

STEP 1

1. We are determining whether certain points are types of discontinuities for given functions.
2. An oscillating discontinuity occurs when a function oscillates infinitely as it approaches a point.
3. An infinite discontinuity occurs when a function approaches infinity as it approaches a point.

STEP 2

1. Analyze the point $x = 0$ for the function $y = \sin \frac{1}{x}$ .
2. Analyze the point $x = \frac{\pi}{2}$ for the function $y = \tan x$ .

STEP 3

Analyze the point $x = 0$ for the function $y = \sin \frac{1}{x}$ :
The function $y = \sin \frac{1}{x}$ is defined for all $x \neq 0$ . As $x$ approaches 0, the value of $\frac{1}{x}$ becomes very large, causing $\sin \frac{1}{x}$ to oscillate between -1 and 1 infinitely. This is characteristic of an oscillating discontinuity.

STEP 4

Analyze the point $x = \frac{\pi}{2}$ for the function $y = \tan x$ :
The function $y = \tan x$ is defined as $\frac{\sin x}{\cos x}$ . At $x = \frac{\pi}{2}$ , $\cos x = 0$ , which makes the function undefined. As $x$ approaches $\frac{\pi}{2}$ from the left, $\tan x$ approaches $+\infty$ , and from the right, it approaches $-\infty$ . This is characteristic of an infinite discontinuity.
The statements are:
1. False: $x = 0$ is indeed an oscillating discontinuity for $y = \sin \frac{1}{x}$ .
2. True: $x = \frac{\pi}{2}$ is an infinite discontinuity for $y = \tan x$ .

Was this helpful?

Get Studdy

Scan QR code with your device to get Studdy on iOS or Android

QR code

Studdy solves anything!

Download the app

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

Parents Influencer program Contact Policy Terms