Limits

**Lotte1990** · March 20th 2012, 06:06 AM

How do I find these limits? Please help!

$\lim_{x \to 14}\frac{\sqrt{x+2}-4}{70-19x+x^2}$

$\lim_{x \to 13}\frac{-26+2x}{\sqrt{x+3}-4}$

**princeps** · March 20th 2012, 06:33 AM

Originally Posted by Lotte1990

How do I find these limits? Please help!

$\lim_{x \to 14}\frac{\sqrt{x+2}-4}{70-19x+x^2}$

$\lim_{x \to 13}\frac{-26+2x}{\sqrt{x+3}-4}$

$L_1=\displaystyle \lim_{x \to 14}\frac {(x-14)}{(x-14)(x-5)(\sqrt{x+2}+4)}=\frac{1}{72}$

$L_2=2\cdot \displaystyle \lim_{x \to 13} \frac{(x-13)(\sqrt{x+3}+4)}{(x-13)}=16$

**Lotte1990** · March 20th 2012, 06:38 AM

Can you please explain your calculation? I need some more information... Thanks!

**Prove It** · March 20th 2012, 06:47 AM

For the first one, rationalise the numerator by multiplying top and bottom by the top's conjugate.
For the second one, rationalise the denominator by multiplying top and bottom by the bottom's conjugate.

**HallsofIvy** · March 20th 2012, 07:02 AM

A useful fact for problems of this sort: if P is a polynomial such that P(a)= 0 then x- a is a factor or P(x).

Also, if you have a term involving a root that is 0 when x= a, then rationalizing it will give a polynomial with a factor of x- a.

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