Understanding the Limit of a Rational Function at a Point

[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 existent 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.

 

[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 :)

 

[HallsofIvy]

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

 

[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.

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

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

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

 

[HallsofIvy]

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

 

[Raza]

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

Sorry :( I won't do that again.

 

[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.

 

[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 $x^2-3^2=(x+3)(x-3)$.