/*! 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} Q. 5 TF Interesting trigonometric limits... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Interesting trigonometric limits: For each of the functions that follow, use a calculator or other graphing utility to examine the graph of f near x=0. Does it appear that f is continuous at x=0? Make sure your calculator is set to radian mode

The function fis continuous atx=c.

Short Answer

Expert verified

The proof islimx→x0(f(x)-f(c))=0

Step by step solution

01

Step 1. Given information

The function fis differentiable at x=cthen the function f is continuous atx=c.

02

Step 2. Calculation

A function is said to be continuous over a range if it is graph is a single and unbroken curve.

Formally,

A real valued function f(x)is said to be continuous at a point x=x0in domain if

limx→x0f(x)exists and is equal to f(x0)

If a function f(x)is continuous at x=x0then

limx→x0+f(x)=limx→x0-f(x)=limx→x0f(x)

Function that are not continuous are said to be discontinuous

Hence proved,

limx→x0f(x)=limx→cf(c)limx→x0f(x)-limx→cf(c)=0limx→x0(f(x)-f(c))=0

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.