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If the value of my answer when I compute the limit of a function as it tends to c is 0. Not infinity. Is it then continuous?

That is the limit as $x$ tends to $90$ in the function $f(x) = \cos (x)$

The answer of the above is 0, is the function continuous at 90? Thanks

Edit: Please determine the continuity or otherwise of $$f(x) = \frac{x-36}{\sqrt x - 36}$$ at the point x = 36.

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You want to say, I suppose, "as $x$ tends to $90^\circ$". Yes, the function is continuous at $x=90^\circ$, since $\lim\limits_{x\rightarrow90^\circ}\cos(x)=0=\cos(90^\circ)$. –  David Mitra Jun 6 '12 at 18:43
    
Do you really mean $$\frac{x-36}{\sqrt{x}-36}\ ?$$ Or do you mean $$\frac{x-36}{\sqrt{x}-6}\ ?$$ –  Arturo Magidin Jun 6 '12 at 18:54
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@Mob: If you add material to the question that substantially changes the question, then please indicate so explicitly. –  Arturo Magidin Jun 6 '12 at 18:56
    
The former. Just how it is in the question. –  Mob Jun 6 '12 at 19:01
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1 Answer

A function $f(x)$ is continuous at $c$ if and only if three things happen:

  1. The function is defined at $c$.
  2. The limit of $f(x)$ as $x\to c$ exists.
  3. The value of the function at $c$ equals the value of the limit: $$\lim_{x\to c}f(x) = f(c).$$

So just from "the limit of a function as it tends to $c$ is $0$" you cannot tell whether the function is continuous or not: you need the value of the function at $c$ to also be $0$ to be able to conclude that.

In the special case of $f(x) = \cos(x)$ at $x=\frac{\pi}{2}$ (measuring in radians):

  1. $f(x)$ is defined at $x=\frac{\pi}{2}$.
  2. The limit of $\cos(x)$ as $x\to\frac{\pi}{2}$ exists: $$\lim_{x\to\frac{\pi}{2}} \cos x = 0.$$
  3. The value of the limit agrees with the value of the function, since $\cos(\frac{\pi}{2}) = 0$. Since $$\lim_{x\to 0}f(x) = f(0),$$ then you can conclude that $f(x)$ is continuous at $0$.

That is: it's not just the numerical value of the limit that matters, it is whether that numerical value agrees with the value of the function at that point that matters.


To the new question: the function $$f(x) = \frac{x-36}{\sqrt{x}-36}.$$

At $x=36$, the function is defined, with value $0$. The limit of the function can be computed using limit laws, since $\sqrt{x}$ is continuous at $36$. We get: $$\begin{align*} \lim_{x\to 36}f(x) &= \lim_{x\to36}\frac{x-36}{\sqrt{x}-36} \\ &= \frac{\lim_{x\to 36}(x-36)}{\lim\limits_{x\to36}(\sqrt{x}-36)}\\ &= \frac{36-36}{\sqrt{36}-36} = \frac{0}{-30} = 0. \end{align*}$$ Since $f(x)$ is defined at $36$, has a limit as $x\to 36$, and the value of the limit equals the value of the function, then the function is continuous at $x=36$.

If you meant $f(x) = \dfrac{x-36}{\sqrt{x}-6}$, then this function is not continuous at $x=36$, because it is not defined at $x=36$.

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@ Arturo. Very nice, +1 –  Rick Decker Jun 6 '12 at 18:51
    
Thanks. Edited the question –  Mob Jun 6 '12 at 18:53
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