# Thread: Limits problems without #'s

1. ## 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.

2. But Dan, we all know that if $f(x)=\sqrt x\implies f'(x)=\frac1{2\sqrt x}$

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

3. Wow! You make it seem so easy! Thanks a bunch!

[EDIT] So wait...then the answer is 2√x?

4. 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!

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

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

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

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

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

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

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

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

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

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

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

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

-Dan

5. Originally Posted by Krizalid
But he's asking for $f(x)=\frac1{\sqrt x}$ which is $f'(x)=2\sqrt x$
Oops!, I made a mistake here , I confused the derivative for an integral

6. Originally Posted by Krizalid
Oops!, I made a mistake here , I confused the derivative for an integral
It's okay. We still love you. (In a platonic way, of course! )

-Dan

7. 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?

8. 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.
Originally Posted by topsquark
$f^{\prime}(x) = -\frac{1}{2}x^{-3/2}$

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

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

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

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

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

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

-Dan

9. Guess it was just a mathematical error on my part. Thanks for all your help today topsquark!

10. Originally Posted by KennyYang
Guess it was just a mathematical error on my part. Thanks for all your help today topsquark!
No problem!

-Dan

11. 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.