Limits & continuity: f = x - 2 for x <= 3, x - 1 for x > 3

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I do not understand some of my math problems. If someone could explain how to figure out the following. Once I see a problem explained I am able to do the rest. Here goes. It may be hard be I can't do all the symbols.

consider

f(x) = x-2 for x less than or equal 3
x-1 for x greater than 3
So ?
Find lim f(x)
x approaches 3+ from the right

OR if you need another problem

Find lim f(x)
x approaches-1

Thanks in advance.
 
Re: Limits & continuity

Are you sure you stated the problem correctly?. You have

\(\displaystyle f(x)=\begin{Bmatrix}x-2, \;\ if \;\ x\leq{3}\\x-1, \;\ if \;\ x<3\end{Bmatrix}\)

Should the top one be \(\displaystyle x\geq{3}\)?.
 
Re: Limits & continuity

I am sorry they bottom one should be GREATER than not less than. The top is correct.
 
Re: Limits & continuity

Since you're approaching f(x) from the right of 3, i.e. dealing with values GREATER than x = 3 ... you can just directly evaluate f(3) using f(x) = x - 2 since you have the theorem:

\(\displaystyle \lim_{x \to a} f(x) = f(a) \quad \mbox{If f(x) is a polynomial or a rational function and a is in its domain}\)
 
Re: Limits & continuity: f = x - 2 for x <= 3, x - 1 for x < 3

I do not know how to work the problem out. The book tells me the answer is 2 but I can get to that conclusion.????
 
Re: Limits & continuity: f = x - 2 for x <= 3, x - 1 for x < 3

Given you're problem ... that's not the right answer.

\(\displaystyle \lim_{x \to 3^{+}} f(x) \quad \mbox{Right?}\)
 
Oh sorry misread which one was the '>'. Well in that case, then instead of using f(x) = x - 2 ... use f(x) = x - 1 since you're considering values for x > 3.
 
Okay NOW I get what I have to do. I didn't even know where to start. That brings me to another question btu I will start of new thread.

THANKS
 
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