- #1

danago

Gold Member

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**Evaluate [tex]\mathop {\lim }\limits_{x \to 3} \frac{{\sqrt {x^2 - 5} - 2}}{{x - 3}}[/tex].**

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:

[tex]\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}[/tex]

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.