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06-20-2009

Since this is a piecewise function you must evaluate
$\lim_{x\rightarrow 3^+}f(x)$
and
$\lim_{x\rightarrow 3^-}f(x)$

Both limits are equal to infinity. (Graph the equation, chances are you will learn why later.)

For the second question:

The way f(x) is defined as so:

$ f(x)= \begin{cases} \sqrt{4-x^2}&-2<x<2\\ 4x-\sin(3.14x)&2\le x \end{cases} $

So for x=2, you need to evaluate $4(2)-\sin(3.14\cdot2)$

If you want to find if the limit of f(x) as x approaches 2, you need to find the limit of f(x) as x approaches 2 from the positive side and the negative side. If these are equal, then the limit exists and is equal to that number. (Of course it is not so the limit does not exist).

Third Question:
$g\circ f(x)=g(f(x))$ . From there, you should see what to do.

Remember the definition of $1^+$ . This means that you evaluate the function from the positive side of 1. Since this function is undefined for $x>1$ (Why?) the limit from the positive side does not exist and therefore, the limit of $\sqrt{1-x}$ at x=1 does not exist.

06-20-2009	#3
hunter34 Guest Posts: n/a	Since this is a piecewise function you must evaluate $\lim_{x\rightarrow 3^+}f(x)$ and $\lim_{x\rightarrow 3^-}f(x)$ Both limits are equal to infinity. (Graph the equation, chances are you will learn why later.) For the second question: The way f(x) is defined as so: $<br /> f(x)=<br /> \begin{cases}<br /> \sqrt{4-x^2}&-2<x<2\\<br /> 4x-\sin(3.14x)&2\le x<br /> \end{cases}<br />$ So for x=2, you need to evaluate $4(2)-\sin(3.14\cdot2)$ If you want to find if the limit of f(x) as x approaches 2, you need to find the limit of f(x) as x approaches 2 from the positive side and the negative side. If these are equal, then the limit exists and is equal to that number. (Of course it is not so the limit does not exist). Third Question: $g\circ f(x)=g(f(x))$ . From there, you should see what to do. Remember the definition of $1^+$ . This means that you evaluate the function from the positive side of 1. Since this function is undefined for $x>1$ (Why?) the limit from the positive side does not exist and therefore, the limit of $\sqrt{1-x}$ at x=1 does not exist.