# Limits

• Aug 30th 2009, 09:36 AM
flutterby
Limits
These are typical examples of the limits I am having trouble on. I need help and I have test on Monday eek I understand most limits I just have trouble with these. =D Thank You-- Carol-ANN

PS. I know how to do number 4
• Aug 30th 2009, 10:40 AM
DeMath
For 1)

$\displaystyle \mathop {\lim }\limits_{\Delta x \to 0} - \frac{{\sqrt {\left( {x + \Delta x} \right) - 8} - \sqrt {x + 8} }}{{\Delta x}} =$

$\displaystyle = - \mathop {\lim }\limits_{\Delta x \to 0} \frac{{\left( {\sqrt {\left( {x + \Delta x} \right) - 8} - \sqrt {x - 8} } \right)\left( {\sqrt {\left( {x + \Delta x} \right) - 8} + \sqrt {x - 8} } \right)}} {{\Delta x\left( {\sqrt {\left( {x + \Delta x} \right) - 8} + \sqrt {x - 8} } \right)}} =$

$\displaystyle = - \mathop {\lim }\limits_{\Delta x \to 0} \frac{{\left( {x + \Delta x} \right) - 8 - \left( {x - 8} \right)}} {{\Delta x\left( {\sqrt {\left( {x + \Delta x} \right) - 8} + \sqrt {x - 8} } \right)}} =$

$\displaystyle = - \mathop {\lim }\limits_{\Delta x \to 0} \frac{1}{{\sqrt {\left( {x + \Delta x} \right) - 8} + \sqrt {x - 8} }} = - \frac{1}{{2\sqrt {x - 8} }}.$
• Aug 30th 2009, 01:34 PM
eXist
For #2)
Substitute and apply Hospital's rule:

$\displaystyle \lim_{x\rightarrow0}\frac{1-cos^2x}{x} = \lim_{x\rightarrow0}\frac{sin^2x}{x} = \lim_{x\rightarrow0}\frac{\frac{d}{dx}(sin^2x)}{\f rac{d}{dx}(x)} = \lim_{x\rightarrow0}\frac{2sinxcosx}{1} =$
$\displaystyle \frac{2sin(0)cos(0)}{1} = 0$

For #3)
Look at this graph:
http://www.intmath.com/Trigonometric-graphs/tanx.gif
As $\displaystyle x\rightarrow\frac{\pi^+}{2}$ you get $\displaystyle -\infty$

For #5)
It's a similar problem. You're approaching form the right (positive side) so substitute 1.1 in. You're going to get a really large negative number. Now imagine as $\displaystyle 1.1\rightarrow1.0$ it is going to approach $\displaystyle -\infty$