Trying To Use Squeeze Theorem To Prove Derivatives Are Equal

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Homework Statement
Problem Statement In Image Below
Relevant Equations
Squeeze Theorem and Limit Definition of Derivative
Below is the question and my attempt at a solution. From the info in the problem I tried to use the squeeze thm to show limf(x)=limg(x)=limh(x) all as x goes to a. Then that combined with f(a)=g(a)=h(a) I used to say all 3 derivatives are equal. Is my attempt below correct or did I make an error somewhere along the way. Thanks in advance.
SQUEEZE 2A.JPG
 
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Without losing generality we can do
a=0, f(a)=g(a)=h(a)=0.
When x>0
\frac{f(x)}{x} \leq \frac{g(x)}{x} \leq \frac{h(x)}{x}
x ##\rightarrow## +0
f'(0) \leq \frac{g(x)}{x} \leq h'(0)
Do similar when x <0
 
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toslowtogofast2a said:
Homework Statement: Problem Statement In Image Below
Relevant Equations: Squeeze Theorem and Limit Definition of Derivative

Below is the question and my attempt at a solution. From the info in the problem I tried to use the squeeze thm to show limf(x)=limg(x)=limh(x) all as x goes to a. Then that combined with f(a)=g(a)=h(a) I used to say all 3 derivatives are equal. Is my attempt below correct or did I make an error somewhere along the way. Thanks in advance.
View attachment 355021
First of all, allow me to invite you to use the LaTeX capabilities of Physics Forums when posting mathematical expressions. Find a link to the LaTeX Guide here or at the end of this Thread. You are likely to get better responses to your posts if you use LaTeX.
 
toslowtogofast2a said:
Homework Statement: Problem Statement In Image Below
Relevant Equations: Squeeze Theorem and Limit Definition of Derivative

Below is the question and my attempt at a solution. From the info in the problem I tried to use the squeeze thm to show limf(x)=limg(x)=limh(x) all as x goes to a. Then that combined with f(a)=g(a)=h(a) I used to say all 3 derivatives are equal. Is my attempt below correct or did I make an error somewhere along the way. Thanks in advance.
View attachment 355021
I would say you haven't proved anything. You've more or less just stated that the result is obvious.
 
PS the question hints at using a direct ##\epsilon-\delta## approach, which might be a good exercise.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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