Studying real analysis on my own. I've come across the following problem. $$f(x) = x^2 + 2x + 6; \text{ show } \lim_{x\to3}f(x)=21$$ The solution that was given was:  Choose $\epsilon&gt;0$:   $|f(x)-21|=|x^2+2x-15|$ ...(so far so good we just subtracted the L value from the function)   $=|x-3||x+5|$   ...(normally I would just set $&lt;ε$ and solve for $|x-3|$, but can't do that here as I would still have a variable term)  Then the solution says, restrict $\delta&lt;1$, ..(where $\delta$ represents the size of the window we consider as $x$ approaches $3$. Here is the part that is troubling me. I don't understand why we can do this) The solution then says: we want $|x-3|&lt;\delta&lt;1$ $$\implies-1&lt;x-3&lt;1$$ $$\implies-&gt;7&lt;x+5&lt;9$$ $$\implies|x+5|&lt;9$$ Then it says: $$|f(x)-21|=|x-3||x+5|&lt;9|x-3|&lt;\epsilon \text{ if } |x-3|&lt;\epsilon/9$$ Take $\delta=\min(1,\epsilon/9)$ $$\text{Then if } |x-3|&lt;\delta\implies|f(x)-2|&lt;\epsilon$$ $$\lim_{x\to3}f(x)=21$$ I don't understand the intuition behind just restricting $\delta&lt;1$. $\delta$ is the window we must be inside to enable us to get arbitrarily close to the limit value. Why do we and how can we restrict ourselves to the case where $\delta&lt;1$? As I don't really understand this point, I don't understand much of the rest. I don't see why we are taking the minimum value either $\delta=\min(1,\epsilon/9)$ Can somebody help?
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I'm trying to understand the proof of the following limit using epsilon-delta definition.  $$\lim _{x\to3} x^2 = 9$$  In Stewart Calculus, the proof goes like this:    I'm confused at the following points:  1) How the author arrives at this result: $|x-3| &lt; \varepsilon/C = \delta$?   Specifically, how,  $$|(x+3)(x-3)| &lt; \varepsilon$$ and $$|(x+3)(x-3)| &lt; C|(x-3)|$$ leads to $$|x-3| &lt; \varepsilon/C = \delta.$$ This is the part I've trouble understanding.  2) What's the role played by $\delta=\min(1, \varepsilon/7)$ ?  I am trying to understand this epsilon-delta proofs for a week but to no avail.
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How can I show:      $$\lim_{x\to\infty}x^{2/3}((x+1)^{1/3}-x^{1/3})=1/3$$   I've tried multiplying with its "conjugate" but that doesn't seem to help that much.  Thanks!
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I'm looking for a rigorous proof of the statement: $\delta(x) = \lim_{\epsilon-&gt;0} \frac{\sin(x/\epsilon)}{\pi x}$ (see ).  For any non-zero value of x, LHS of the above is by definition zero. But, for any non-zero value of x, the limit in RHS simply does not exist. So, how is this statement proved?
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Be $f:\mathbb R^ +\mapsto\mathbb R$ a function that satisfies the following conditions:  a)$ f(f(f(x)))+2x=f(3x)$ for every $x\gt 0$;  b)  $\lim_{x \to \infty} (f(x)-x)=0$.  This was proposed by Gabriel Dospinescu  at a romanian competitions about a decade ago. No idea how to approach it. Thank you!
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This was the question of a test. My question is if my attempt to solve it is correct, and if it is, why is it correct.  $$\lim_{(x,y) \to (0,0), x+y \neq 0}{\frac{\ln(1-x-y)}{x+y} } $$  My attempt:   Let $\xi = -x-y $. Then $\xi \to 0$ whenever $(x,y) \to (0,0)$ and $x+y \neq 0 \iff \xi \neq 0$. (Is it then correct to say that the previous limit exists and is equal to the following iff the following exists? And why?):  $$\lim_{\xi \to 0, \xi \neq 0}{\frac{\ln(1+\xi )}{-\xi}}$$   If it is correct, then the limit exists and is $-1$. If it is correct, why is it correct?
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I know the question is a weird title.     Define $f:\mathbb{R}\to\mathbb{R}$ by   $$\begin{cases}     8x &amp; \text{if $x$ is rational}\\     2x^2+8 &amp; \text{if $x$ is irrational}    \end{cases}$$           Determine at which points of $\mathbb{R}$ the function $f$ has a limit and justify your conclusions.   So we see that it is possible that they may approach the same point at $x=2$. Since this is an accumulation point for both parts of the function we can inquire as to the existence of a limit here. We don't consider other points since there will clearly be no other accumulation points that might give the same limit.  I'm confused on how to show the limit formally exists though I am pretty sure it converges to 16 obviously. So I know I have something of the form $0&lt;|x-2|&lt;\delta$ and $|f(x)-16|&lt;\epsilon$. Where I am stuck is talking about the function itself. I don't think showing this with sequences will work but I'm not sure.  Thanks!
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My Calculus book asks us to evaluate the limit:  $$ \lim_{n \to \infty} \sum_{i=1}^{n} f(c_{i}) \Delta x_{i}  $$  of f(x) = $\sqrt{x}$ within [0,3]  Why is the right endpoint given as $3\frac{i^2}{n^2}$?  Similarly, $c_{i}$ becomes $\frac{i^3}{n^3}$ for $\sqrt[3]{x}$ within [0,1].  I apologize if my LaTeX is bad, my first attempt.  It doesn't explain it very well in the book.  Thanks!
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Question:  Prove that $\lim _{x \to \infty} x\sin x$ doesn't exist (using delta epsilon).   What I did: I've been struggling with this one for a long time. Really tried digging up the net for explanations, but couldn't find a really good source to explain how this proof should be done(some said proof by contradiction like I did, and some say without contradiction). I got too confused by now, I also read this  which gave some guidance but I thought I'd use the floor function to make sure M is natural, otherwise the idea of using these values wouldn't have worked. Anyway I'd really love someone to verify/correct this solution before I hand it in. Thanks a lot.   We assume there exists a limit, therefore:  $\forall \varepsilon&gt;0 \exists M&gt;0: x&gt;M \Rightarrow |f(x)-L|&lt;\varepsilon$ and specifically, if we pick some $x_1,x_2&gt;M$: $|f(x_1)-L|&lt;\frac \varepsilon2,  |f(x_2)-L|&lt;\frac \varepsilon2$ Also, this applies for all $\varepsilon$, and specifically for  $\varepsilon=0.5$ (1) Therefore using the triangle inequality: $|f(x_1)-f(x_2)| = |f(x_1)-L+L-f(x_2)| \le |f(x_1)-L |  + |f(x_2)-L| \le \frac \varepsilon2 +  \frac \varepsilon2 = \varepsilon$ This applies for all $x&gt;M$ so specifically for: $x_1=\frac {\pi \lfloor M+1 \rfloor}{2} &gt; M , x_2=\pi \lfloor M+1 \rfloor$ We put these values in (1): $|f(x_1)-f(x_2)|=|\frac {\pi \lfloor M+1 \rfloor}{2}\sin\frac {\pi \lfloor M+1 \rfloor}{2}-\pi \lfloor M+1 \rfloor \sin (\pi \lfloor M+1 \rfloor)|= \begin{cases} 1 &amp; \lfloor M+1 \rfloor   odd  \\ 0 &amp;\lfloor M+1 \rfloor  even\end {cases} $ So we have found an $\varepsilon $ that doesn't always satisfy the requirements of the limit definition so this is a contradiction. Therefore the limit doesn't exist.
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Prove that $\lim_{x\to 2}x^2=4$ using $\epsilon-\delta$ definition.   By the mean of $\epsilon-\delta$ definition, $|x-2|\le \delta,|x+2|\le \delta+4$  then $|x-2||x+2|\le \delta(\delta+4),|x^2-4|\le \delta^2+4\delta$.  Assign $\epsilon=\delta^2+4\delta,|x^2-4|\le\epsilon$.  Q.E.D.  Is this method correct? If yes, why I always see people do this by putting $\delta = \min(1,\epsilon/5)$...?  Thanks.
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I was wondering what is the value of $\lim\limits_{x\to 0+} x^{x}$.  Assuming that the limit exists, I could show using the usual logarithm techniques that the limit is $1$. However, I am not able to show that the limit exists. Could some one help on that?  EDIT: It would be so nice if we could do it without l'Hopital...
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How would I go about proving this limit?  $\lim_{x\to\infty}\frac{1}{x^4+x^2+5}=0$  so far I have:  $|f(x) -L|&lt; ϵ$ wherever $x &gt; N$  $|\frac{1}{x^4+x^2+5} - 0| &lt; ϵ $ wherever $x &gt; N$  $|\frac{1}{x^4+x^2+5}| &lt; ϵ  ,\; x &gt; ∞$, assuming, $x &gt; 0$  taking the absolute value $\frac{1}{x^4+x^2+5} &lt; ϵ$  ${x^4+x^2+5}$ > $\frac{1}{ϵ}$  i am not sure whether this is correct, but what would be the next step?
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Show that $|x|&lt;1$ implies that $\lim_{n \to \infty}x^n$ = 0.   Can anyone help me define my $N$ (as used in the formal definition of a limit) in terms of $\varepsilon$? I define $x$ as $\dfrac{1}{1+p}$ where $p&gt;0$ but I am not sure how also to define $N$ since both $P$ and $\varepsilon$ are variable.
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I am trying to prove,  $\lim_{x \to a} \sqrt x = \sqrt a$  As per the definition of limit for every $\epsilon &gt; 0$ there is some $\delta &gt; 0$ such that   $ 0 &lt; |x-a| &lt; \delta$ implies $|f(x) - \sqrt a| &lt; \epsilon$  $|\sqrt x - \sqrt a|  = \frac {|x -a|}{\sqrt x + \sqrt a} $  since  $ 0 &lt; |x-a| &lt; \delta$  $\frac {|x -a|}{\sqrt x + \sqrt a} &lt; \delta$  Can I stop my proof at this point since I found $\epsilon$ ( which is $\delta$ in this case)
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I'm starting  and finally decided to learn how to write epsilon-delta proofs.  I have been working on chapter 5, number 3(ii). The problem, in essence, asks to prove that     $$\lim\limits_{x \to 1} \frac{1}{x} = 1.$$    Here's how I started my proof,     $$\left| f(x)-l \right|=\left| \frac{1}{x} - 1 \right| =\left| \frac{1}{x} \right| \left| x - 1\right| &lt; \epsilon \implies \left| x-1 \right| &lt; \epsilon |x|$$   I haven't made any further progress past this point. Is it possible to salvage this proof? Should I try an alternate approach?
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Hello I was hoping that someone might be able to verify that the following proof that $\lim_{x\to 0} {1\over x}$ does not exist is correct.   First assume that $\lim_{x\to 0} {1\over x} = L$. This means that for every $\epsilon &gt; 0$ there is a $\delta &gt; 0$ such that for all $x$ if $0 &lt; \lvert x\rvert &lt; \delta$ then $\lvert{1\over x} - L \rvert &lt;\epsilon.$  Now let $\epsilon=1$ and choose $x &lt; \min(1,\delta)$.  Therefore there is a $\delta&gt;0$ such that if $0 &lt; \lvert x\rvert &lt; \delta$ then $\lvert{1\over x} - L \rvert &lt;\epsilon$.  It follows that $\lvert{1\over x}\rvert &lt; 1+ \lvert L\rvert$.  However clearly $\lvert{1\over x}\rvert&gt;1$ and therefore a contradiction has been reached and there is no number $L$ such that $\lim_{x\to 0} {1\over x} = L$.
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I have been in a deep confusion for about a month over the topic of limits! According to our book, the limit at $a$ is the value being approached by a function $f(x)$ as $x$ approaches $a$.  I have a doubt that in real number line we can never ever reach the closest value to $a$ because always a more closer value will exist.  Now when talking about our methods for calculating limit, what method comes in our mind when we have to calculate where the value approaches? Let's say we have to find what value $f(x)$ approaches when $x$ moves from $[0,a)$.  So we calculate the value of $f(x)$ at $x=0$, say $4$, and then at $x\to a$.   But the problem is that we don't know the value $x\to a$, so we say the value is $a-h$ where $h\to 0$ and calculate the value of $f(a-h)$, say $5-h$.  Now here is where my doubt starts! In the final step we put the value of $h=0$ say in that it is an infinitesimal quantity.  My doubt is that $h$ was tending to $0$ means that it was never equal to zero  maybe it is infinitesimally small, not a stationary value, not an imaginable value, but we know for sure that it is not  equal to $0$. Maybe it is the point closest to zero, but it is not equal to zero and when we use the result $5-h=5$ we are actually making an error which is tending to zero. Maybe the error is very small, but still, there is some error in that we cannot calculate it but we can see that there is this infinitesimal error present.  That means we don't get the exact limiting value or last value of $f(x)$.  $x$ belongs to $[0,a)$ but a value approximate to infinitesimal? Isn't it right! We get an approximated value?
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Using my right to ask for help once again. I can't solve the following problem for quite a long time. Just have no idea how to do that. Here it is. Let $f(x)$ be bounded on any interval $(1,b), b&gt;1$. Then $\lim_{x\to+\infty}{f(x) \over x}=\lim_{x\to+\infty}(f(x+1)-f(x))$. It is from the Russian book "Lekcii po matematicheskomu analizu" by G.I.Arkhipov, V.A.Sadovnichy and V.N.Chubarikov, 2nd ed., Moscow, 2000, page 676. Please, keep in mind, that times, when I did homework had passed long time ago. Now I'm solving problems for pleasure.