Can anyone help me finish this proof?  Prove that $$\lim_{x\to 2} \left(x^3\right) = 8$$  $|x^3-8|=|(x-2)(x^2+2x+4)|=|x-2|&lt;1$ $$1&lt;x&lt;3$$
123144
I am concern about the delta calculation given $0 &lt; x &lt; 4$. I believe it works, but I am not 100 percent sure as I am new to proofs.  Consider the function $f(x) = x^3$ for $x ∈ ℝ$ and $0 &lt; x &lt; 4$. Prove that $$\lim _{ x\rightarrow 2 }{ { x }^{ 3 }=8 } $$  Proof: Let $\varepsilon  &gt; 0$ be given, then let $\delta =  \min   {(\varepsilon/19,1)}$. Now suppose $|x -2| &lt; \delta = \min {(\varepsilon /19,1)}$. Then $|x^3 - 8| = |x - 2||x^2 + 2x + 4|&lt; |x - 2|&lt;19$  $(|x^2 + 2x + 4| &lt; 19~\text{if}~|x - 2| &lt; 1)&lt; \varepsilon /19 $  $(|x - 2| &lt; \varepsilon /19 )= \varepsilon $.  Thus this $\delta$ makes $|x^3 - 8| &lt; \varepsilon $ whenever $0 &lt; |x - 2| &lt; \delta$. Therefore, it follows that $\lim _{ x\rightarrow 2 } x^3 = 8$.
122318
Let $f\left(x\right)=x^{13}+3x^{9}+2x^{5}+x^{2}+x+2$ find $\left(f^{-1}\right)'\left(1\right)$. Can somebody show me the method? I am a little confused.
98609
How would I prove that if $f: \mathbb R^2 \to \mathbb R$ is a function such that  $$\lim_{(x,y)\to(a,b)} f(x,y) = L$$  and for every $y_0 \in \mathbb R$   $$ \lim_{x\to a} f(x,y_0) = L'_{y_0}$$  and for every $x_0 \in \mathbb R$   $$ \lim_{y\to b} f(x_0,y) = L''_{x_0}$$  then  $$ \lim_{x\to a}\left(\lim_{y\to b} f(x,y)\right)  = \lim_{y\to b}\left(\lim_{x\to a} f(x,y)\right) = L$$
88464
Does $\lim_{x \to 0+} \left(x\lfloor \frac{a}{x} \rfloor\right)=a?$   I'm going to say this statement is false, and try to use the properties of limits  $$\lim_{x \to 0+} \left(x\lfloor \frac{a}{x} \rfloor\right)=\lim_{x \to 0+} x.\lim_{x \to 0+}\lfloor \frac{a}{x}\rfloor$$ =$0.\infty$ which is undefined.  Or is it $\infty $ because x ends up being a little bigger than 0?
106369
Prove that for $x&gt;0$  and $0&lt;p&lt;q$:  $$\left( 1+ \frac{x}{p} \right) ^{p}  &lt; \left( 1+ \frac{x}{q} \right)  ^{q} $$
88599
Find   $$\lim _{x\to \infty }x\Big(\ln\left(x+1\right)-\ln x\Big)$$  Here's how I do it:  $$x\Big(\ln(x+1)-\ln x\Big) = x\Bigg(\ln(x(1+\frac{1}{x}) - \ln x\Bigg)$$ $$x\Big(\ln x + \ln(1+\frac{1}{x}\Big)-\ln x =x\ln\Big(1+\frac{1}{x}\Big)$$ $$\ln\Bigg( \Big(1+\frac{1}{x}\Big)^x \Bigg) \rightarrow \ln 1^\infty = 0$$  What am I doing wrong? The answer is supposed to be $1$, but I get $0$.
16469
I have seen a proof that shows $$\lim_{n\to\infty} \left(1+\frac{x}{n}\right)^n = e^x$$ by looking at the Taylor series expansion of $\ln(1+x)$ at $x=0$. To prove a theorem, my textbook uses the fact $$\lim_{n \to \infty} \left(1+\frac{a_n}{n}\right)^n = e^a$$ when $a_n \to a$. How can I prove this?
31850
Show that $$n! \leqslant \left(\frac{n+1}{2}\right)^n \quad \hbox{for all } n \in \mathbb{N}$$  I know that it can be done by induction but I always find line where I do not know what to do next.
16328
I need help solving this problem. I tried L'hospital and rearranging but nothing worked. $$ \lim_\limits{x→∞} x^2\left(\ln\left(1 + \frac{1}{x}\right)- \frac{1}{x+1}\right) $$
42270
Give an $\varepsilon-\delta$ proof using the “$\delta\leq 1 $ trick” to prove that $\lim_{x\rightarrow c} (2x^ 2 − 3x + 4) = 2c ^2 − 3c + 4$.   My attempt: Let $ε&gt;0$ and there exist $\delta&gt;0$ such that whenever $|x-c|&lt;δ$,  $$|2x^ 2 − 3x + 4-(2c ^2 − 3c + 4)|&lt;ε$$  $$\implies |2x^2-2c^2-3x+3c|&lt;ε$$  $$\implies 2|x^2-c^2|-3|x-c|&lt;ε$$  $$\implies 2|x-c||x+c|-3|x-c|&lt;ε$$  $$\implies |x-c||2x+2c-3|&lt;ε$$  I don't  see where to go from here.
87671
How to use the formal definition of the limit of a sequence to prove the following limit?  $\lim_{x \to -2} {(x^3 + 4x^2+ 4x-1)} = -1,$  $\lim_{x\to 1}{(\frac{1}{x})} = 1.$
18978
Here's the question: Prove that $\lim_{n \to \infty} (\sqrt{n^2+n}-n) = \frac{1}{2}.$  Here's my attempt at a solution, but for some reason, the $N$ that I arrive at is incorrect (I ran a computer program to test my solution against some test cases, and it spits out an error). Can anyone spot the error for me?  $\left|\sqrt{n^2+n}-n-\frac{1}{2}\right| &lt; \epsilon$  $\Rightarrow \left|\frac{n}{\sqrt{n^2+n}+n} - \frac{1}{2}\right| &lt; \epsilon$  $\Rightarrow \frac{1}{2}-\frac{1}{\sqrt{1+\frac{1}{n}}+1} &lt; \epsilon$  $\Rightarrow \frac{1}{\sqrt{1+\frac{1}{n}}+1} &gt; \frac{1}{2} - \epsilon = \frac{1-2 \epsilon}{2}$  $\Rightarrow \frac{1}{\sqrt{1+\frac{1}{n}}} &gt; \frac{1-2 \epsilon}{2}$  $\Rightarrow \frac{1}{\sqrt{\frac{1}{n}}} &gt; \frac{1-2 \epsilon}{2}$  $\Rightarrow \sqrt{n} &gt; \frac{1-2 \epsilon}{2}$  $\Rightarrow n &gt; \frac{4 {\epsilon}^2-4 \epsilon +1}{4}$
6534
I need to prove this limit:     Given $f:(-1,1) \to \mathbb{R}\,$ and $\,f(x)&gt;0,\,$ if $\,\lim_{x\to 0} \left(f(x) + \dfrac{1}{f(x)}\right) = 2,\,$ then $\,\lim_{x\to 0} f(x) = 1$.
47565
I get that the limit is  $$\lim\limits_{n\to\infty}f(x)=1$$ because $$\lim\limits_{n\to\infty}(1+x^{2n})=1,\,\,\lvert x\rvert&lt;1.$$  Is this right?
22753
Prove that if one of $\displaystyle\lim_{x \rightarrow 0 }{f(x)}$ and $\displaystyle\lim_{x \rightarrow 0}{f(x^3)}$ exists, then the other one also exists. Can anyone guide me on this ? I have no idea on how to start .
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.
38305
Assume that $x_n$ is a sequence such that $x_n$ converge to a constant $x$. I would like to prove that   $\left( 1+ x_n/n \right)^n \rightarrow e^x$  Obvious if $x_n = x$ the result would be given. The extra sequence complicates things and I would like a hint or solution on how to proof this.