Continuity of function with parameter

[Aaron17]

I have this function. I need to find all the parameters a (real number) so that the function is continuous.



When I substitute first I get a limit of type 0/0. If I extend the expression by a/a and make adjustments I get as a result a. For the second limit I get a/3 + a. So I get a = a / 3 + a which is probably not good. How should I work with the first limit?

In the end a got a = a/3 + a. So parameter a is equal to 0?

 

[tkhunny]

When you substitute what you get "a limit type of 0/0"? Are there ANY continuity problems for ANY value of "a" when you are nowhere near x = 1?

Your ONLY concern is AT x = 1. Is it continuous, there? If you do not ALREADY know it's continuous, you shouldn't be substituting anything. Find any value of "a" that MAKES it continuous.

You should also know that [math]\lim_{x\rightarrow 0 }\dfrac{\sin(a\cdot x)}{a\cdot x} = 1[/math]

 

[HallsofIvy]

Not a very interesting problem, is it?
Yes, we can write \(\displaystyle \frac{sin(ax- a)}{x- 1}\) as \(\displaystyle \frac{a}{a}\frac{sin(ax- a)}{ax- a}\) and, as a goes to 1, ax-a goes to 0. So letting y= ax- a the limit ,as x goes to 1 from below, is the same as \(\displaystyle \lim_{y\to 0}\frac{sin(y)}{y}\) which is well known to be 1.

So \(\displaystyle \lim_{x\to1}\frac{sin(ax-a)}{x- 1}= a\)

Of course, \(\displaystyle \lim_{x\to 1}\frac{a}{x+2}+ ax= \frac{a}{3}+ a\).

In order that the function be continuous those two limits must be the same: a= a/3+ a so a/3= 0, a= 0.

Which means that the function is identically 0, 0 for all x!

As I said, not very interesting.