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Okay. So I solved this question, but I'm not sure its correct.

It is a two-part question and goes like this:

 

a) Determine $$\lim_{h\rightarrow0} \frac{(5+h)^{-1} -5^{-1}}{h}$$

 

b) Interpret the answer using the word "slope", and without the use of the word "limit"

 

-------------------------------

 

For the first part of the question I simplified the equation and direct substituted the zero, thus getting 0 as an answer. 

 

For the second part I said, "The slope at the function at x=5 is indeterminate." 

 

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 If you need me to, I can scan in the paper I did the work on, to see how I solved the problem. Is what I did correct or complete nonsense?

chilledz3non  Sep 19, 2014

Best Answer 

 #10
avatar+91481 
+11

Dragon,  please only comment when you can be helpful.  You know chilledz was not talking to you and that time you had nothing helpful to say!  

Your gradient post was okay,  Chilledz might have found that helpful. 

 

 

Yes, it is the slope OF the tangent line at x=5.   

Melody  Sep 20, 2014
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11+0 Answers

 #1
avatar+91481 
+11

Are you sure that the question is correct?

$$\\\lim_{h\rightarrow0} \frac{(5+h)^{-1} -5^{-1}}{h}\\\\
=\lim_{h\rightarrow0} \frac{\frac{1}{5+h}-\frac{1}{5}}{h}\\\\
=\lim_{h\rightarrow0} \left(\frac{5}{5(5+h)}-\frac{5+h}{5(5+h)}\right)\div h\\\\
=\lim_{h\rightarrow0} \left(\frac{-h}{5(5+h)}}\right)\times \frac{1}{h}\\\\
=\lim_{h\rightarrow0}\; \frac{-1}{5(5+h)}}\\\\
=\frac{-1}{5(5+0)}}\\\\
=\frac{-1}{25}}\\\\$$

$$The gradient of $ y=\frac{1}{x} $ at $ x=5 $ is $ \frac{-1}{25}$$

Melody  Sep 20, 2014
 #2
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Thanks for answering Melody. Yes, the question is written exactly like how it is on the paper. I just texted someone in my class. We think that it is just a bullsh#$ question.... -_- 

 

What do you think? Is this a trick question?

chilledz3non  Sep 20, 2014
 #3
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No it is my mistake  -  give me  a little while and I will fix it.

Melody  Sep 20, 2014
 #4
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+11

ok I fixed it.   

Melody  Sep 20, 2014
 #5
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Oh okay! Thanks so much Melody! I was sure the professor was pulling my leg, he did it once before lol.

Also what is gradient? Is that another word for function?

chilledz3non  Sep 20, 2014
 #6
avatar+91481 
+11

Gradient is rise over run the same as when you talk about the gradient of a line.

The gradient of a curve is really the gradient of the tangent to the curve at a given point.

A tangent is a line that just touches the curve.    :)

Melody  Sep 20, 2014
 #7
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Take a look.

DragonSlayer554  Sep 20, 2014
 #8
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Oh okay, so its the slope at the tangent line?

chilledz3non  Sep 20, 2014
 #10
avatar+91481 
+11
Best Answer

Dragon,  please only comment when you can be helpful.  You know chilledz was not talking to you and that time you had nothing helpful to say!  

Your gradient post was okay,  Chilledz might have found that helpful. 

 

 

Yes, it is the slope OF the tangent line at x=5.   

Melody  Sep 20, 2014
 #11
avatar+564 
+6

Ah okay, thank you! :)

chilledz3non  Sep 20, 2014
 #13
avatar+91481 
+6

This site might help

http://www.teacherschoice.com.au/first_principles.htm

 

I am trying to find an interactive site or calculator to demonstrate this - I am sure I have seen one before. 

But I can't find it.    

Melody  Sep 20, 2014

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