Im sorry if this is stupid or obvious, but why   $$ e=\lim_{n\rightarrow \infty}\left(1+\frac{1}{n}\right)^n$$  AND  $e$ is the unique positive number for which $$ \lim_{h\rightarrow 0}\frac{e^h-1}{h}=1$$?  I mean, how do we know that these two definitions are equivalent?   Again, maybe it's easy, but I'm just beginning calculus and our teacher just dropped those two definitions...
124045
Show that, $$e=\lim_{x\to \infty} \left(1+\frac{1}{x}\right)^x $$  Is the same number that satisfies,  $$\lim_{h\to0} \frac{e^{h}-1}{h} = 1 \tag{*}$$  You don't have to do that if it's too cumbersome, but (*) is used to find the derivatives of real exponential functions. I know that definitions are not to be proven but I'm looking for some derivation or intution for the claim that (*) is a legitimate definition for e, and furthermore, a proof that limit in (*) exists.
98810
Self learning Analysis and found the following exercise in Bartle's Elements of Real Analysis:     Let $X = (x_n)$ be a sequence of strictly positive real numbers such   that $\lim \left({\frac {x_{n + 1}} {x_n}} \right) \lt 1.$Show that   for some $r$ with $0\lt r \lt 1$ and some $C \gt 0 $ we have $ \ 0 \lt  x_n \lt Cr^n$ for all sufficiently large $n \in \Bbb N$. Use this to   show that $\lim(x_n) = 0.$   Now I haven't proven any significant results concerning sequences and hence am required to solve this problem with basically nothing but the definition of a limit.   I am not getting too far with this because I cannot see a way to get there with a direct implication and I cannot for the life of me negate "for all sufficiently large $n$" in the context of the sentence above.   Would be extremely grateful if someone could help me out.  What I've done so far:  We know that $l = \lim \left({\frac {x_{n + 1}} {x_n}} \right) \lt 1$. If $l \lt 0$ we can find a neighbourhood of $l$ entirely contained in $(-\infty, 0)$ which would contain no elements of the sequence $  \left({\frac {x_{n + 1}} {x_n}} \right) $ all of whose elements are positive leading to a contradiction. This implies $l \ge 0.$  We can pick $ 0 \lt  \epsilon \lt 1 - l$. There is $m \in \Bbb N$ such that   $$n \ge m \implies \left|{\frac {x_{n + 1}} {x_n} - l}\right| \lt \epsilon \implies \frac {x_{n + 1}} {x_n} \lt \epsilon + l \lt 1 \implies x_{n + 1} \lt x_n$$      Also can someone tell me if $\lim (x_n) = 0$ necessarily? Doesn't   $\left({2 + \frac 1 n}\right) $ satisfy the given conditions??
83864
In fact the real question is: Does $\lim\limits_{n\to\infty}1^{n}=e$?.  I know that   $$ \lim\limits_{n\to\infty}\left(1+\dfrac{1}{n}\right)^n=e, $$  So, can we say that $1^\infty=e$?  And, by logic, This product  $$\underset{\text{infinity times}}{1\cdot1\cdot\ldots\cdot1},$$ gives $1$ not other value.
90860
We know that $\lim_{n\rightarrow\infty}(1+\frac{1}{n})^n=e$ and so I thought the approach to evaluating the limit in the question would be to just use this fact and substitute it into the numerator. This approach would tell us the above limit evaluates to $1$. However, that does not seem to be the correct limiting value. In fact, it evaluates to $\frac {1}{\sqrt{e}}$. Why is this so?
107616
Why is it that   $$\lim_{n \rightarrow +\infty} \left( 1 + \frac{\lambda}{n}+ o \left( \frac{1}{n} \right) \right)^n = e^{\lambda}$$  I once remembered the justification for "ignoring" the $o(1/n)$ but can't recall it again.
52861
Finding $\displaystyle \lim_{y\rightarrow \infty}\bigg[\frac{y}{e}-y\bigg(\frac{y}{y+1}\bigg)^y\bigg]$   Attempt: Put $\displaystyle y=\frac{1}{z}$ . Then $\displaystyle \lim_{z\rightarrow 0}\bigg[\frac{1}{ze}-\frac{1}{z}\bigg(\frac{z}{z+1}\bigg)^{\frac{1}{z}}\bigg]$  Could some help me to solve it, thanks
78063
Find $$\lim_{n \rightarrow \infty}\left(\int_0^1(f(x))^n\,\mathrm dx\right)^\frac{1}{n}$$if $f:[0,1]\rightarrow(0,\infty)$ is a continuous function.   My attempt:  Say $f(x)$ has a max. value $M$. Then $$\left(\int_0^1(f(x))^ndx\right)^\frac{1}{n}\leq\left(\int_0^1M^ndx\right)^\frac{1}{n} =M$$  I cannot figure out what to do next.
86681
Find $\lim\limits_{n\rightarrow \infty} \frac{n^t}{e^n}$ where $t$ is a positive real number.   My answer is that the expression is indeterminate ($\frac{\infty}{\infty}$) but it will tend to $0$ because as $n$ gets large, the exponential term will dominate the polynonial term in the expression, so the expression will tend to $0$ as $n$ gets large. This answer is also the same as the solution provided in my text.  But I am wondering what is the actual workings to solve this? I'd know to because I want to be familiar with all the tricks to solve limits where the expression is indeterminate.
48528
I want to prove the following limit:  $$\lim_{n \to \infty} \sqrt[n+1\;] {(n+1)!} - \sqrt[n] {(n)!} = \frac{1}{e}.$$  I searched the forum &amp; found the link here:  .  But still, there is no way out of the problem. So, please solve it.
16963
How to prove that the following definitions of $e^x$ are equivalent, with as simple tools as possible and without any knowledge of $e$ or logarithms ?  $$\sum_{n=0}^{\infty} \frac{x^n}{n!}=\lim_{n\to\infty}\left(1+\frac{x}{n}\right)^n$$  Also preferably, prove that this is $a^x$ for some real number $a&gt;0$.
58845
$\lim_{n\to\infty}h_n(x) = x\lim_{n\to\infty}x^{\frac{1}{2n-1}}$ where $h_n(x)=x^{1+\frac{1}{2n-1}}$. I understand that $\lim_{n\to\infty}x^{\frac{1}{2n-1}}$ goes to one but what I don't understand is how did our limit become $h_n(x)=|x|$? I'm just having hard time wrapping my head around the appearance of absolute value.  Note. This is an example (Chapter 6, Section 2) from Understanding Analysis by Abbott.
72656
While I was reading the book Exploratory examples for real analysis, I came across this:     to show that : $$\lim_{x\rightarrow x_o}\frac{1}{h(x)} = \frac{1}{\lim_{x\rightarrow x_o}{h(x)}} = \frac{1}{M}$$      Discussion:Let $\epsilon \gt 0$.we want to find $\delta \gt 0$ such that       if  $0 \lt |x-x_0| \lt \delta $,then $\Big|\frac{1}{h(x)}-\frac{1}{M}\Big|\lt \epsilon$.      In order to get expression involving $|h(x)-M|$,we first find the common denominator:        $\Big|\frac{1}{h(x)}-\frac{1}{M}\Big|=\Big|\frac{M-h(x)}{M~h(x)}\Big|$      $~=\Big|\frac{h(x)-M}{|M|~|h(x)|}\Big|$.      As we have to deal with variable term $|h(x)|$ in denominator.We can construct bound .Let $\epsilon =\frac{|M|}{2}$ ,since lim$_{x\to x_o}h(x)=M$,there exists $\delta_1\gt 0~~$ s.t.       if $0 \lt |x-x_0| \lt \delta_1 $,then $|h(x)-{M}|\lt \frac{|M|}{2}$.      As you study steps below,identify that step where $h(x)\neq 0$ for all $0 \lt |x-x_0| \lt \delta_1 $ plays a critical role:      $|M|=|M-h(x)+h(x)|$      $~~~~~~\leq |M-h(x)|+|h(x)|$      $~~~~~~= |h(x)-M|+|h(x)|$      $~~~~~~\lt \frac{M}{2}+|h(x)| \implies$      $\frac{|M|}{2} \lt |h(x)| \implies$      $\frac{1}{h(x)}\lt \frac{2}{|M|}$,where $0 \lt |x-x_0| \lt \delta_1 $.      Now that we've bound for $1/h(x)$,we can apply the definition to term $|h(x)-M|$.      Since     lim$_{x\rightarrow x_o}h(x)=M$,we can find $\delta_2 \gt 0$ such that,       $~~~~~~~~~~~~~~~~~~~~~~~~~~~~$if  $0 \lt |x-x_0| \lt \delta_2$, then $|h(x)-M|\lt \frac{|M|^2 \epsilon }{2}$   The thing I can't understand is for what purpose did the author work with such complicated expression involving $\epsilon $ in the end of the discussion?
79693
I think it is zero; $$\lim_{n \rightarrow +\infty } \left(\sqrt[m]{\prod_{i=1}^{m}(n+{a}_{i})}-n\right)$$ we can make that steps: $$\lim_{n \rightarrow +\infty } \left(\sqrt[m]{{n}^{m}\prod_{i=1}^{m}\left(1+\frac{a_i}n\right)}-n\right)$$ and then: $$\lim_{n \rightarrow +\infty } \left(n \cdot \sqrt[m]{\prod_{i=1}^{m}\left(1+\frac{{a}_{i}}{n}\right)}-n\right)$$ each bracket aspires to unit so we have:  $$\lim_{n\rightarrow \infty} (n-n)$$ and..  $$\lim_{n\rightarrow \infty} (0) = 0$$
6138
We have $$e^{2\pi i n}=1$$  So we have $$e^{2\pi in+1}=e$$  which implies $$(e^{2\pi in+1})^{2\pi in+1}=e^{2\pi in+1}=e$$ Thus we have $$e^{-4\pi^{2}n^{2}+4\pi in+1}=e$$  This implies $$e^{-4\pi^{2}n^{2}}=1$$  Taking the limit when $n\rightarrow \infty$ gives $0=1$.
97560
Let's analyze this expression   $\lim_\limits{n\rightarrow\infty} (1+\frac{1}{n})^n$  It's the definition of $e$ which, as we know is not equal to $1$. So what is wrong with the following "logic":  As $\lim_\limits{n\rightarrow\infty}(a_{n}b_{n}) = \lim_\limits{n\rightarrow\infty}(a_{n})\times\lim_\limits{n\rightarrow\infty}(b_{n}) $ and $\lim_\limits{n\rightarrow\infty}(1+\frac{1}{n}) = 1$, we can say that $\lim_\limits{n\rightarrow\infty} (1+\frac{1}{n})^n=1^n$, which is equal to $1$.  I know something's wrong there, but the question is - what?
97123
I'm very eager to know and understand the definition of $e$. Textbooks define $e$ as follows $$ e = \lim_{p\to\infty} \left[1+\frac1{p}\right]^p \approx 2.71828 $$ Is there an &quot;easy to understand&quot; proof of this? I'm really looking for a derivation of this which is very intuitive and easy to comprehend. By the way I'm watching .
14865
Prove that if $u_n\rightarrow \infty$ then $\frac{1}{u_n} \rightarrow 0$  Could someone please point me in the right direction? I don't know how to show this at all. Any help would be much appreciated :)