MA129 Lecture Notes - Lecture 15: Classification Of Discontinuities, Continuous Function

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16 Feb 2019

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MA129 Lecture Notes - Lecture 15: Classification Of Discontinuities, Continuous Function

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= l f (x) lim x a as y l. This means that the limit of x a be real numbers. L a y = f (x) (x) as be a function of approaches x y = f. February 8, 2019 x is equal to a. L (dr. allison) x > 1 x < 1 f f (x) (x) decreases to 4. increases to 4. Example: lim x 1 x + 3 = 4 from values of from values of x 1 x 1. Formally: one sided limits lim x 1+ x + 3 = 4 lim x 1 . February 8, 2019 lim x 2 x + 2 x x 6. Find the limit, if the limit exists: 0, 1. 1: 5, the limit does not exist. Cross out common multiples in the numerator and denominator. X = 2 lim x 2 x + 2 x x 6. Note: there is a hole due to a removable discontinuity at.

Related Questions

Hello,

Theproblem asks:
For what values of constants a andb is the function f(x) continous on (-infinity, + infinity) ?

x + 2 , x < 1
f(x) = a , x = 1
bx^2 , x > 1

If you have time, please read below and confirm or correct mythinking on this problem:

To me the problem is saying:

That as the limit of x approaches 1 from the left(1-) and/or from the right (1+); f(x) =3.

So, for the limit to be continuous, f(x) = 3 it will haveto be true [when x is/close to 1]for each of the three (piecewiseequations? what is the correct terminology?).

So I think that means a = 3 and b = 3 when x = 1 or xapproaches 1. To check this mathematically, do I solve for eachvariable? I.e. 3 = bx^2 and 3 = a?

Am I understanding this problem correctly? If you havetime, please explain (in writing) what this problem means and howto solve it.

Thanks!

ceruleanchimpanzee878

MA129 Lecture Notes - Lecture 15: Classification Of Discontinuities, Continuous Function

MA129 Lecture 13: Matrices - Adding, Subtracting, Multiplying

MA129 Lecture Notes - Lecture 15: Classification Of Discontinuities, Continuous Function

MA129 Lecture 16: Limits, Continuity, and Rates of Change

Document Summary

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