# Limits problems without #'s

• Aug 14th 2007, 11:44 AM
KennyYang
Limits problems without #'s
Does this problem make anyones else's eyes bug out?

If f(x) = 1/(√x) find lim [f(x+h)-f(x)]/ h
h->0

b. If the above limit represents a slope function for the line tangent to f at (x, f(x)). Write the equation of the tangent line at x=4, in point-slope form.
• Aug 14th 2007, 12:53 PM
Krizalid
But Dan, we all know that if $\displaystyle f(x)=\sqrt x\implies f'(x)=\frac1{2\sqrt x}$

But he's asking for $\displaystyle f(x)=\frac1{\sqrt x}$ which is $\displaystyle f'(x)=2\sqrt x$
• Aug 14th 2007, 12:54 PM
KennyYang
Wow! You make it seem so easy! Thanks a bunch!

[EDIT] So wait...then the answer is 2√x?
• Aug 14th 2007, 01:09 PM
topsquark
Quote:

Originally Posted by KennyYang
Does this problem make anyones else's eyes bug out?

If f(x) = 1/(√x) find lim [f(x+h)-f(x)]/ h
h->0

b. If the above limit represents a slope function for the line tangent to f at (x, f(x)). Write the equation of the tangent line at x=4, in point-slope form.

Krizalid isn't quite right here.

Round 2! :)

$\displaystyle f(x) = \frac{1}{\sqrt{x}}$

$\displaystyle f^{\prime}(x) = \lim_{h \to 0} \frac{\frac{1}{\sqrt{x + h}} - \frac{1}{\sqrt{x}}}{h}$

Simplify a bit first:
$\displaystyle f^{\prime}(x) = \lim_{h \to 0} \frac{\frac{1}{\sqrt{x + h}} - \frac{1}{\sqrt{x}}}{h} \cdot \frac{\sqrt{x + h} \sqrt{x} }{\sqrt{x + h} \sqrt{x} }$

$\displaystyle = \lim_{h \to 0} \frac{\sqrt{x} - \sqrt{x + h}}{h\sqrt{x + h} \sqrt{x}}$

Again, rationalize the numerator:
$\displaystyle = \lim_{h \to 0} \frac{\sqrt{x} - \sqrt{x + h}}{h\sqrt{x + h} \sqrt{x}} \cdot \frac{\sqrt{x} + \sqrt{x + h}}{\sqrt{x} + \sqrt{x + h}}$

$\displaystyle = \lim_{h \to 0} \frac{x - (x + h)}{h\sqrt{x + h} \sqrt{x}(\sqrt{x} + \sqrt{x + h})}$

$\displaystyle = \lim_{h \to 0} \frac{-h}{h\sqrt{x + h} \sqrt{x}(\sqrt{x} + \sqrt{x + h})}$

$\displaystyle = -\lim_{h \to 0} \frac{1}{\sqrt{x + h} \sqrt{x}(\sqrt{x} + \sqrt{x + h})}$

$\displaystyle = -\frac{1}{\sqrt{x} \sqrt{x}(\sqrt{x} + \sqrt{x})}$

$\displaystyle = -\frac{1}{x(2\sqrt{x})}$

$\displaystyle = -\frac{1}{2x \sqrt{x}}$

Which is typically written as $\displaystyle -\frac{1}{2}x^{-3/2}$

-Dan
• Aug 14th 2007, 01:12 PM
Krizalid
Quote:

Originally Posted by Krizalid
But he's asking for $\displaystyle f(x)=\frac1{\sqrt x}$ which is $\displaystyle f'(x)=2\sqrt x$

Oops!, I made a mistake here :eek:, I confused the derivative for an integral :D:D
• Aug 14th 2007, 01:23 PM
topsquark
Quote:

Originally Posted by Krizalid
Oops!, I made a mistake here :eek:, I confused the derivative for an integral :D:D

It's okay. We still love you. (In a platonic way, of course! :eek: )

-Dan
• Aug 14th 2007, 02:14 PM
KennyYang
I gave part (b) a shot and came up with such a convoluted answer that I'm not going to bother to post it. Did anyone else get a ridiculously overcomplicated answer as well?
• Aug 14th 2007, 02:42 PM
topsquark
Quote:

Originally Posted by KennyYang
b. If the above limit represents a slope function for the line tangent to f at (x, f(x)). Write the equation of the tangent line at x=4, in point-slope form.

Quote:

Originally Posted by topsquark
$\displaystyle f^{\prime}(x) = -\frac{1}{2}x^{-3/2}$

-Dan

Any line can be expressed in the form $\displaystyle y = mx + b$. The slope of the tangent line at a point (x, f(x)) of the graph is $\displaystyle f^{\prime}(x)$, so the slope of the tangent line is, in this case:
$\displaystyle m = f^{\prime}(4) = -\frac{1}{2 \cdot 4^{3/2}} = - \frac{1}{2 \cdot 8} = -\frac{1}{16}$

So we have the form:
$\displaystyle y = -\frac{1}{16}x + b$

Now, we know that the function f(x) and the tangent line have the same values at the point x = 4. So we know that
$\displaystyle f(4) = y(4) = -\frac{1}{16} \cdot 4 + b$

$\displaystyle f(4) = \frac{1}{\sqrt{4}} = \frac{1}{2} = -\frac{1}{16} \cdot 4 + b$

$\displaystyle b = \frac{1}{2} + \frac{1}{16} \cdot 4 = \frac{3}{4}$

So I get that the tangent line to f(x) at x = 4 is $\displaystyle y = -\frac{1}{16}x + \frac{3}{4}$.

-Dan
• Aug 14th 2007, 03:15 PM
KennyYang
Guess it was just a mathematical error on my part. Thanks for all your help today topsquark!
• Aug 15th 2007, 07:41 AM
topsquark
Quote:

Originally Posted by KennyYang
Guess it was just a mathematical error on my part. Thanks for all your help today topsquark!

No problem! :)

-Dan
• Aug 16th 2007, 04:18 AM
curvature
Quote:

Originally Posted by KennyYang
If f(x) = 1/(√x) find lim [f(x+h)-f(x)]/ h
h->0

In fact the limit is the derivative of f(x) at x.