# Basic Limits Question

1. Mar 26, 2007

### danago

Evaluate $$\mathop {\lim }\limits_{x \to 3} \frac{{\sqrt {x^2 - 5} - 2}}{{x - 3}}$$.

I started by trying direct substitution, which gave me a solution of an indeterminate form (0/0). I then used L'hopitals rule, and evaluated using derivatives:

$$\begin{array}{l} \mathop {\lim }\limits_{x \to 3} \frac{{\sqrt {x^2 - 5} - 2}}{{x - 3}} = \frac{0}{0} \\ \mathop {\lim }\limits_{x \to 3} \frac{{\sqrt {x^2 - 5} - 2}}{{x - 3}} = \mathop {\lim }\limits_{x \to 3} \frac{x}{{\sqrt {x^2 - 5} }} = \frac{3}{2} \\ \end{array}$$

I then went to check my answer by making a table on my calculator. From that, it seemed that my right hand limit was negative infinity, while my left hand limit was infinity. Since each side of the limit is different, the limit is non existant at x=3.

Which one of my solutions is correct (if any), and why is the other wrong? This has caused a bit of confusion.

Thanks in advance,
Dan.

2. Mar 26, 2007

### danago

Hmm nevermind. I checked again on my calculator, and it seems i entered my function incorrectly. My new, correct, function yields the answer i was looking for.

Thanks anyway :)

3. Mar 26, 2007

### HallsofIvy

Staff Emeritus
Far simpler than "L'Hopital" is to multiply both numerator and denominator of the fraction by $\sqrt{x^2-5}+ 2$.

4. Mar 27, 2007

### Raza

I might be wrong

$$\mathop {\lim }\limits_{x \to 3} \frac{{\sqrt {x^2 - 5} - 2}}{{x - 3}}$$

$$\mathop {\lim }\limits_{x \to 3} \frac{{\sqrt {(3)^2 - 5} - 2}}{{(3) - 3}}=\frac{0}{0}$$

$$\mathop {\lim }\limits_{x \to 3} \frac{{\sqrt {x^2 - 5} - 2}}{{x - 3}}\times\frac{{\sqrt {x^2 - 5} + 2}}{{\sqrt {x^2 - 5} + 2}}$$

$$\mathop {\lim }\limits_{x \to 3} \frac{{x^2 - 5 - 4}}{{(\sqrt {x^2 - 5}+2)(x-3)}}$$

$$\mathop {\lim }\limits_{x \to 3} \frac{{(x-3)(x+3)}}{{(\sqrt {x^2 - 5}+2)(x-3)}}$$

The (x-3)'s get crossed out.

$$\mathop {\lim }\limits_{x \to 3} \frac{{x+3}}{{\sqrt {x^2 - 5}+2}}$$

$$\mathop {\lim }\limits_{x \to 3} \frac{{(3)+3}}{{\sqrt {(3)^2 - 5}+2}}$$

$$\mathop {\lim }\limits_{x \to 3} \frac{{6}}{{4}}}=\frac{3}{2}$$

Last edited: Mar 27, 2007
5. Mar 28, 2007

### HallsofIvy

Staff Emeritus
Raza, please do not solve the complete problem! Leave something for the original poster, whose homework this is, to do.

6. Apr 19, 2007

### Raza

Sorry :( I won't do that again.

7. Apr 19, 2007

### denverdoc

just wondering how in the 4th step from the bottom, it got factored into:
(x-3)(x+3), trying to buff math skills and missed that one completely.

8. Apr 20, 2007

### Gib Z

Because $x^2-5-4=x^2-9=x^2-3^2$. By the difference of two squares rule, $(a-b)(a+b)=a^2-b^2$, we attain $x^2-3^2=(x+3)(x-3)$.

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