December 12th, 2016, 05:55 PM  #1 
Newbie Joined: Dec 2016 From: Sweden Posts: 13 Thanks: 1  Hard limit problem one variable calculus
Made errors in earlier posts. This is my question. I have been trying to solve this since summer. It is a question from the book: 7th edition Adams calculus. 2.1.3 Using Newton quotient to solve the problem (not any chain rule, power rule or l'HÃ´pital's rule is allowed). If the function $f$ has a tangent at the given point? If yes, what is the tangent line? $\displaystyle f(x)=\sqrt{x}\: \text{at} \: x=0$ It looks obvious at first: $\displaystyle \lim_{h\to 0}\frac{\sqrt{0+h}\sqrt{0}}{h}=$ $\displaystyle =\:\lim_{h\to 0}\frac{\sqrt{h}}{h}$ Which does seem right. I can draw the needed conclusion I think? It is a twosided limit, which does not exist in this case. But when I look in the book. The solutions manual. It states the same conclusion as me, but with a different way of reaching it. It says: $\displaystyle \lim_{h\to 0}\frac{\sqrt{0+h}\sqrt{0}}{h}=$ Up until this point I understand, but then the solutions manual Bends My mind Into deep insanity with the above claiming to be equal to $\displaystyle =\lim_{h\to 0}\frac{1}{h\text{sgn}(h)}$ Can someone please explain to me how this last step is achieved? Is it just a logical step I am supposed to make to show an equal situation as above? Where the limit goes to negative and positive infinity? Or is there really a way to rewrite $\displaystyle \lim_{h\to 0}\frac{\sqrt{0+h}\sqrt{0}}{h}=$ into $\displaystyle \lim_{h\to 0}\frac{1}{h\text{sgn}(h)}$ ?????? I would really love an answer. Thank you. This last step is taken from The instructor's solutions manual. I am assuming it is completely correct. Last edited by skipjack; December 13th, 2016 at 06:07 AM. 
December 13th, 2016, 01:44 AM  #2 
Global Moderator Joined: Dec 2006 Posts: 20,969 Thanks: 2222 
The function $f(x)$ is defined, but not differentiable, at $x = 0$. Its graph has a tangent (with equation $x = 0$) at $x = 0$. Your use of the difference quotient in your post is correct. The solutions manual, if you have quoted it accurately, contains a mistake (a typographical error or slip by the author). 
December 13th, 2016, 03:20 AM  #3 
Math Team Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 
It does not help you to post the same question repeatedly!

December 13th, 2016, 04:55 AM  #4 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,683 Thanks: 2664 Math Focus: Mainly analysis and algebra 
1) Your book is wrong. 2) $\DeclareMathOperator{\sgn}{sgn} h=h \sgn{(x)}$ by definition. 3) The tangent would be the $y$ axis. Last edited by v8archie; December 13th, 2016 at 04:58 AM. 
December 13th, 2016, 06:09 AM  #5 
Global Moderator Joined: Dec 2006 Posts: 20,969 Thanks: 2222 
I take it you mean the $f(x)$ axis, as there is no $y$ in this problem. I've merged the other thread into this one. 
December 13th, 2016, 01:48 PM  #6  
Newbie Joined: Dec 2016 From: Sweden Posts: 13 Thanks: 1  Quote:
Certain? It is not some type of text proof like situation where something is similar to the real answer and I should understand it? That I should use this hsgn(h) as a replacement for a situation where things look the same close to 0? It made me feel stupid. An answer to this would be helpful. I have been retaking calclulus for years. And this is the only one on the first half of the cource I cannot solve. Except for epsilon delta proofs.  
December 13th, 2016, 02:53 PM  #7 
Senior Member Joined: Sep 2016 From: USA Posts: 645 Thanks: 408 Math Focus: Dynamical systems, analytic function theory, numerics 
Maybe the comments above relate to the other thread but the quote you have from the book is not wrong. It is exactly correct. For $h > 0$ you have $f(h) = \sqrt{h}$ and therefore \[ \lim_{h \rightarrow 0^+} \frac{f(0+h)  f(0)}{h} = \lim_{h \rightarrow 0^+} \frac{1}{\sqrt{h}} = \infty. \] Note that if $h < 0$ you have $f(h) = \sqrt{h}$ and a similar computation gives you a limit of $\infty$. This should make sense graphically since the slopes of $f(x)$ increase without bound as $x$ tends to 0 from the right and they decrease without bound from the left. Refer to the graph to see this geometrically. WolframAlpha: Computational Knowledge Engine It is also not true that there is a vertical tangent line at the $y$axis. In fact there is a cusp here and tangency loses meaning. 
December 13th, 2016, 03:24 PM  #8  
Math Team Joined: Dec 2013 From: Colombia Posts: 7,683 Thanks: 2664 Math Focus: Mainly analysis and algebra  Quote:
This is true (hence the "would") in my post. However, if you were to define a tangent at $x=0$ it would be the $y$axis. Last edited by skipjack; December 13th, 2016 at 06:05 PM.  
December 13th, 2016, 04:45 PM  #9 
Newbie Joined: Dec 2016 From: Sweden Posts: 13 Thanks: 1 
My problem is still with understanding The rewriting of; $\displaystyle \lim_{h\to 0}\frac{\sqrt{0+h}\sqrt{0}}{h}=$ into $\displaystyle \lim_{h\to 0}\frac{1}{h\text{sgn}(h)}$ In one variable calculus. It has no tangent. I know some measurement theory. Did a research level crash course one summer; measurement theory. There was some mentioning of pointy peaks being differentiable. But no! It has no tangent! There are other situations where an infinite limit could imply this. But this is not it. Please help me understand the rewriting of this. Last edited by skipjack; December 13th, 2016 at 06:04 PM. 
December 13th, 2016, 05:44 PM  #10 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,683 Thanks: 2664 Math Focus: Mainly analysis and algebra 
I can't see any analytic way of writing the limit as they have done without first accepting that the limit is $\pm \infty$ and then recognising that any positive power of $h$ gives the same result (i.e unbounded  the direction may be different). The decision to write $\DeclareMathOperator{\sgn}{sgn}x \sgn{(x)}$ instead of $x$ is frankly bizarre. Last edited by v8archie; December 13th, 2016 at 05:47 PM. 

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abs, calculus, hard, limit, newton, point, problem, quotient, variable 
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