How to determine the x values where a function is continuous

[TheRedDevil18]

How would I find the x values for which a function is continuous ?, and how to tell whether it is a removable discontinuity, a jump discontinuity, or an infinite discontinuity ?

Suppose the function is sqrt(9-x^2)/x^2-1

 

[ndjokovic]

$\frac{\sqrt{9-x^2}}{x^2-1}=\frac{\sqrt{9-x^2}}{(x-1)(x+1)}$. So your function is not defined on the points 1 and -1, but it is continuous on the other points as a composition of basic continuous functions. To study the behavior of the function near the singularities, you calculate the limits, you will find $$\lim_{x\rightarrow 1} \frac{\sqrt{9-x^2}}{x^2-1}=+ \infty$$ from the right and $-\infty$ from the left. The opposite thing happens for -1.

 

[TheRedDevil18]

$\frac{\sqrt{9-x^2}}{x^2-1}=\frac{\sqrt{9-x^2}}{(x-1)(x+1)}$. So your function is not defined on the points 1 and -1, but it is continuous on the other points as a composition of basic continuous functions. To study the behavior of the function near the singularities, you calculate the limits, you will find $$\lim_{x\rightarrow 1} \frac{\sqrt{9-x^2}}{x^2-1}=+ \infty$$ from the right and $-\infty$ from the left. The opposite thing happens for -1.

Wouldn't the numerator also have a restriction ?, the square root can't be negative. So would it be correct in saying that the graph is continuous on the intervals [-3,-1) U [3,1) ?

 

[arildno]

f is continuous WHEREVER it is DEFINED (i.e, has a function value).

 

[Mark44]

Wouldn't the numerator also have a restriction ?, the square root can't be negative. So would it be correct in saying that the graph is continuous on the intervals [-3,-1) U [3,1) ?

Interval notation goes left to right, so this would be written as [-3, -1) U (1, 3[/color]].

 

[Mark44]

$\frac{\sqrt{9-x^2}}{x^2-1}=\frac{\sqrt{9-x^2}}{(x-1)(x+1)}$. So your function is not defined on the points 1 and -1, but it is continuous on the other points as a composition of basic continuous functions. To study the behavior of the function near the singularities, you calculate the limits, you will find $$\lim_{x\rightarrow 1} \frac{\sqrt{9-x^2}}{x^2-1}=+ \infty$$ from the right and $-\infty$ from the left. The opposite thing happens for -1.

Technically, $\lim_{x\to 1} \frac{\sqrt{9-x^2}}{x^2-1}$ does not exist, since the left and right limits are different (which is what you said). The notation that is often used is
$\lim_{x \to 1^+} \frac{\sqrt{9-x^2}}{x^2-1} = \infty$ and $\lim_{x \to 1^-} \frac{\sqrt{9-x^2}}{x^2-1} = -\infty$.

 

[arildno]

The function is perfectly continuous at (-1,1) as well.