A great refresher. I'm actually working on those tonight, since in series and sequences the limits come back!! I was hoping they would disappear after the derivative
Could I just check that my answer to this question is correct? Well, can I check that the d in my answer is correct at any rate!
Give an e d proof that the function
f(x) = (x^2 + 1)/(x + 1)
is continuous at x = 1.
I got d=min(1, 1.5e)
Thanks!
Delta-Epsilon proofs well-past my BC calc days when I was taking an introductory proofs class. I assume that many individuals going into or emerging of Calc just do not have evidences down sufficient yet to be able to whip one out.
To show $\displaystyle \displaystyle \begin{align*} \frac{x^2 + 1}{x + 1} \end{align*}$ is continuous at $\displaystyle \displaystyle \begin{align*} x = 1 \end{align*}$, you need to show $\displaystyle \displaystyle \begin{align*} \lim_{x \to 1}\frac{x^2 + 1}{x + 1} = 1 \end{align*}$, i.e. that $\displaystyle \displaystyle \begin{align*} |x - 1| < \delta \implies \left| \frac{x^2 + 1}{x + 1} - 1 \right| < \epsilon \end{align*}$. Working on the second inequality we have
$\displaystyle \displaystyle \begin{align*} \left| \frac{x^2 + 1}{x + 1} - 1 \right| &< \epsilon \\ \left| \frac{x^2 + 1}{x + 1} - \frac{x + 1}{x + 1} \right| &< \epsilon \\ \left| \frac{x^2 - x}{x + 1} \right| &< \epsilon \\ \left|x - 1\right| \left| \frac{x}{x + 1} \right| &< \epsilon \\ \left|x - 1\right| \left| 1 - \frac{1}{x + 1} \right| &< \epsilon \end{align*}$
It would now help us if we had an upper bound on $\displaystyle \displaystyle \begin{align*} \left| 1 - \frac{1}{x + 1} \right| \end{align*}$, to give us the maximum value of $\displaystyle \displaystyle \begin{align*} \left| x - 1 \right| \left| 1 - \frac{1}{x + 1} \right| \end{align*}$. If we bound $\displaystyle \displaystyle \begin{align*} |x - 1| < 1 \end{align*}$ (say), that gives
$\displaystyle \displaystyle \begin{align*} |x - 1| &< 1 \\ -1 < x - 1 &< 1 \\ 0 < x &< 2 \\ 1 < x + 1 &< 3 \\ \frac{1}{3} < \frac{1}{x + 1} &< 1 \\ \frac{1}{|x + 1|} &< 1 \end{align*}$
and to get an upper bound on $\displaystyle \displaystyle \begin{align*} \left| 1 - \frac{1}{x + 1} \right| \end{align*}$ we can use the Triangle Inequality
$\displaystyle \displaystyle \begin{align*} \left| 1 - \frac{1}{x + 1} \right| &\leq |1| + \left| -\frac{1}{x + 1} \right| \\ &= 1 + \frac{1}{|x + 1|} \\ &\leq 1 + 1 \\ &= 2 \end{align*}$
So that gives
$\displaystyle \displaystyle \begin{align*} \left| x - 1 \right| \left| 1 - \frac{1}{x + 1} \right| &< \epsilon \\ 2|x - 1| &< \epsilon \\ |x - 1| &< \frac{\epsilon}{2} \end{align*}$
Therefore choosing $\displaystyle \displaystyle \begin{align*} \delta = \min\left\{ 1, \frac{\epsilon}{2} \right\} \end{align*}$ and reversing the process will finish the proof.
To show that $\displaystyle \displaystyle \begin{align*} e^{|x|} + 1 \end{align*}$ is continuous at $\displaystyle \displaystyle \begin{align*} x = 0 \end{align*}$, you need to show $\displaystyle \displaystyle \begin{align*} |x - 0| < \delta \implies \left| e^{|x|} + 1 - 2 \right| < \epsilon \end{align*}$. Working on the second inequality we have
$\displaystyle \displaystyle \begin{align*} \left| e^{|x|} - 1 \right| &< \epsilon \\ \left| e^{|x|} \right| - |1| &< \epsilon \textrm{ since } \left| e^{|x|} \right| - |1| \leq \left| e^{|x|} - 1 \right| \textrm{ by the Reverse Triangle Inequality } \\ e^{|x|} - 1 &< \epsilon \\ e^{|x|} &< \epsilon + 1 \\ |x| &< \ln{\left( \epsilon + 1 \right)} \end{align*}$
So choosing $\displaystyle \displaystyle \begin{align*} \delta = \ln{\left( \epsilon + 1 \right)} \end{align*}$ and reversing the process will complete the proof.
Hello! I hope its right to ask here, since i wanted to use a example from the pdf file, I am just using the proof of example 3 to show my problem.
I almost understand every single step to make the proof, the mainproblem is this step:
|x^2 - 25| = |x + 5||x - 5| < 11|x-5|
I know to get the 11|x-5|, but i dont get why it is less than my original function? In fact, its one of the first times i really work with inequallities, could someone explain me this little simple step? i solved |x+5| for 11, and thats why i can replace |x+5| with 11 which gives me 11|x-5|, but why is this new function < than my old function? arent both functions the same until i put my delta in?
I was taking an introductory proofs class. I think that most people going into or coming out of Calc just don't have proofs down enough yet to be able to whip one out.Luxury Goods Store
I understand that a little bit but I am having a hard time with how to start my proof.
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thanks for the nice paper! so in example 3 do you choose delta=1 arbitrarily and M is equal to the upperbound of 11? can you explain further what the meaning of the M derivation is?
for instance if you choose in ex 3 delta=0.1 instead of 1 then would get M=10.1 and define delta as min{0.1, 10.1} and is that an equivalent proof?
do most mathematicians learn this very quickly?
The idea is that as long as your x is sufficiently close to the c value, then you are guaranteed for your function to be within your tolerance of $\displaystyle \displaystyle \epsilon $. So yes, the value $\displaystyle \displaystyle \delta = 1 $ was chosen arbitrarily. Any number smaller than this will also work as we know this value works.
A proof of a formula on limits based on the epsilon-delta definition. An example is the following proof that every linear function f(x)=ax+b (a,b in R,a!=0) is continuous at every point x_0. The claim to be shown is that for every epsilon>0 there is a delta>0 such that whenever |x-x_0|<delta, then |f(x)-f(x_0)|<epsilon fap turbo 2