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The question is this.

In $h(x) = \dfrac{\sqrt{x+9}-3}{x}$, show that $\displaystyle \lim_{x \to 0} \ h(x) = \frac{1}{6}$, but that $h(0)$ is undefinied.

In my opinion if I use this expression $\displaystyle \lim_{x \to 0} \dfrac{\sqrt{x+9}-3}{x}$ above with the $-3$ inside the square root I got an undefinied expression, but if I put the $-3$ out of the square and I use this expression to calculate the limit $\displaystyle \lim_{x \to 0} \dfrac{\sqrt{x+9}-3}{x}$ I will get $\frac{1}{6}$.

Hereenter image description here a print screen of the original question.

If needed i can post the Pdf of the homework.

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Multiply on top and bottom by $\sqrt{x+9}+3$ and re-evaluate. – Wintermute Mar 28 '13 at 23:54
In this case when I said put the -3 out, I just wanna to say that I´m redefining the function above. – Vinicius L. Beserra Mar 28 '13 at 23:57
The question was edited from the OP's original one. I have edited it back. @ViniciusL.Beserra please make sure this represents your intent. – apnorton Mar 28 '13 at 23:58
The second function is a completely different function from the first function. Recheck the problem statement to ensure you have stated the initial problem correctly, and didn't misunderstand. Perhaps there is a typo, because, yes, the limit of the stated problem is undefined. that's not the case with the alteration. – amWhy Mar 29 '13 at 0:00
1-you can not take $-3$ outside the squareroot as you wish $2$- The limit of the original expression is defintely not $1/6$. – Seyhmus Güngören Mar 29 '13 at 0:04

4 Answers 4

To show something slightly different than the other answers: $$\begin{align} \lim_{x \to 0} \frac{\sqrt{x + 9}-3}{x} &= \lim_{x\to 0}\frac{\sqrt{x + 9}-3}{x}\frac{\sqrt{x + 9}+3}{\sqrt{x + 9}+3} \\&= \lim_{x\to 0} \frac{x+9-9}{x(\sqrt{x + 9}+3)} \\&= \lim_{x\to 0}\frac{1}{\sqrt{x+9}+3} \\ &= \frac{1}{6}. \end{align} $$ Clearly the function isn't defined at $0$ because then you are dividing by $0$.

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There is an $x$ missing after the $3$ in the second row. But this way might be easier if one is not familiar with the rule of L'Hospital. – cthl Mar 29 '13 at 0:46
I think the braces in the second row were not there when I said the $x$ was missing. Now it's okay. – cthl Mar 29 '13 at 11:42
Why is the first expression discontinuous at $x=0$, but the last expression is continuous at $x=0$ (despite the fact that they're algebraically equivalent? – alexqwx Sep 20 '14 at 20:23

As I stated in the comments above, and as demonstrated by anorton:

$$h(x) = \frac{\sqrt{x + 9} - 3}{x} \neq \frac{\sqrt{x+ 9 - 3}}{x} = \frac{\sqrt{x+6}}{x}$$

So ... Assuming the problem statement (from the original source) actually reads:

In $h(x) = \dfrac{\sqrt{x+9}-3}{x}$, show that $\displaystyle \lim_{x \to 0} \ h(x) = \frac{1}{6}$, but that $h(0)$ is undefined.

  • We can take the $\lim_{x\to 0} h(x)\;$ by applying L'Hopital's Rule because, as is, the limit evaluates to the indeterminate form $\frac 00$.
  • If you're not familiar with L'Hopital, then you can multiply the numerator and denominator by the conjugate of the numerator: the conjugate of $\sqrt{x + 9} - 3$ is simply $\sqrt{x + 9} + 3.\;$ Then simplify and evaluate the limit of the resulting expression. (See $(1)$ below.)

Either route taken will give you an expression for which the limit can be evaluated. And as you note: $$\lim_{x\to 0} h(x) = \dfrac 16$$

Now, why is $h(0)$ undefined? Because division by zero is undefined. Period.

But to say that the function is undefined when evaluated at $0$, i.e., that $h(0)$ is undefined is not to say that $\lim_{x \to 0} h(x)$ is undefined. Limits often exist at points for which a function is not defined. After all, the limit as $x \to a$ of $f(x)$ gives you the value to which $f(x)$ approaches as $x \to a$ , irregardless of whether the function is undefined AT $x = a$.

$(1)$ $$ \lim_{x \to 0} \frac{\sqrt{x + 9}-3}{x} = \lim_{x\to 0}\frac{\sqrt{x + 9} - 3}{x}\cdot \frac{\sqrt{x + 9} + 3}{\sqrt{x + 9} + 3}$$ $$= \lim_{x\to 0} \frac{x + 9 - 9}{x(\sqrt{x + 9} + 3)} = \lim_{x\to 0} \frac{x}{x\left(\sqrt{x + 9} + 3\right)} $$ $$ = \lim_{x\to 0}\frac{1}{\sqrt{ x + 9} + 3}= \frac{1}{6}$$

ADDED: There is clearly a typo in the original pdf of the question (the image of which you have now posted.) That is $$\lim_{x\to 0} \frac{\sqrt{x+9 - 3}}{x} = \lim_{x\to 0}\frac{\sqrt{x + 6}}{x} \to \frac{\sqrt{6}}{0}, \text{ so that} \lim_{x\to 0} h(x)\to\infty.$$

So I suspect that was intended for this problem (without the misprint) is the function highlighted above, whose limit, as you observe, is then $1/6$. In either case, though, $h(0)$ is undefined, because we cannot divide by zero.

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You cannot pull the negative 3 out of the square root. For example: $$\sqrt{1-3} = \sqrt{2}i \ne \sqrt{1} - 3 = -2$$

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Of course $h(0)$ is undefined as you cannot divide by zero. As you said $\displaystyle\lim_{x\rightarrow 0}\frac{\sqrt{x+9}-3}x=\frac 1 6$, according to L'Hospital's rule. Why do you care about $\displaystyle\frac{\sqrt{x+9-3}}x\ne h(x)$?

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Yes. Clearly the original problem from the text had the $-3$ not under the radical sign. The OP mistook that "placing it in or out" as the "reason" $h(0)$ is undefined (missing the point of zero division). +1 – coffeemath Mar 29 '13 at 0:14

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