# Solve algebraically: $\lim\limits_{x \to 3} \frac{3-x}{5-\sqrt{x^2+16}}$

$$\lim\limits_{x \to 3} \frac{3-x}{5-\sqrt{x^2+16}}$$

The professor says we can't use l'hopital's rule and must solve algebraically.

• Multiply top and bottom by $5+\sqrt{x^2+16}$. Then factor the resulting polynomial downstairs and cancel what you can. – David Mitra Feb 28 '14 at 15:17

## 2 Answers

Try multiplying the numerator and denominator by the conjugate of the denominator:

Multiply by $$\dfrac{5+ \sqrt{x^2 +16}}{5+\sqrt{x^2 + 16}}$$

$$\lim\limits_{x \to 3} \frac{3-x}{5-\sqrt{x^2+16}} \cdot \dfrac{5+ \sqrt{x^2 +16}}{5+\sqrt{x^2 + 16}} = \lim_{x\to 3}\dfrac{(3-x)(5 + \sqrt {x^2 + 16)}}{25 - (x^2 + 16)}$$

Then note that you have a difference of squares in the denominator: $$25 - (x^2 + 16) = 9 - x^2 = (3-x)(3+x)$$

Now, you can cancel the factor $3-x$, as it appears in both the numerator and denominator.

• Thank you so much for you quick reply! I understand and found the limit of 5/3! – dev Feb 28 '14 at 15:28
• Exactly! Nice work, dev. – Namaste Feb 28 '14 at 15:28

Hint $\ \ f\bar f = (a\!-\!x)(a\!+\!x)\ \Rightarrow\ \dfrac{a-x}f \,=\, \dfrac{\bar f}{a+x}\ \$ where $\ \ \bar f,\,f\, =\, 5\pm \sqrt{x^2+16}$

Remark $\$ This is a special case of rationalizing the denominator, which is often handy.

• Does that link imply those three accounts all belong to you? – TMM Feb 28 '14 at 15:49