Linked Questions

6
votes
5answers
746 views

Evaluate $\lim_{x\to 0} \frac{a^x -1}{x}$ without applying L'Hopital's Rule. [duplicate]

The questions is: Evaluate $$\lim_{x\to 0} \frac{a^x -1}{x}$$ without applying L'Hopital's Rule. Does this question fundamentally same as asking if the $\lim_{x\to 0} \frac{a^x -1}{x}$ exists? rather ...
6
votes
3answers
154 views

Elementary proof that $2^x$ is derivable [duplicate]

I was wondering if there was an elementary proof, so not using the exponential function, that $2^x$ is derivable. I define the function $f(x) = 2^x$ by $f(a/b) = \sqrt[b]{2^a}$ for a and b integers, ...
437
votes
25answers
137k views

How to prove that $\lim\limits_{x\to0}\frac{\sin x}x=1$?

How can one prove the statement $$\lim_{x\to 0}\frac{\sin x}x=1$$ without using the Taylor series of $\sin$, $\cos$ and $\tan$? Best would be a geometrical solution. This is homework. In my math ...
13
votes
6answers
15k views

Proving $(1 + 1/n)^{n+1} \gt e$

I'm trying to prove that $$ \left(1 + \frac{1}{n}\right)^{n+1} > e $$ It seems that the definition of $e$ is going to be important here but I can't work out what to do with the limit in the ...
10
votes
5answers
996 views

The derivative of $e^x$ using the definition of derivative as a limit and the definition $e^x = \lim_{n\to\infty}(1+x/n)^n$, without L'Hôpital's rule

Let's define $$ e^x := \lim_{n\to\infty}\left(1+\frac{x} {n}\right)^n, \forall x\in\Bbb R $$ and $$ \frac{d} {dx} f(x) := \lim_{\Delta x\to0} \frac{f(x+\Delta x) - f(x)} {\Delta x} $$ Prove that $...
10
votes
3answers
983 views

If $f$ is continuous and $f(x+y)=f(x)f(y)$, then $\lim\limits_{x \rightarrow 0} \frac{f(x)-f(0)}{x}$ exists

I'm solving the functional equation $f(x+y)=f(x)f(y)$ and I know that I have a continuous function $f:[0,\infty\rangle \to \langle 0,\infty\rangle$ s.t. $f(0)=1$. In one of the steps, I want to show ...
16
votes
3answers
1k views

Can $\lim_{h\to 0}\frac{b^h - 1}{h}$ be solved without circular reasoning?

In many places I have read that $$\lim_{h\to 0}\frac{b^h - 1}{h}$$ is by definition $\ln(b)$. Does that mean that this is unsolvable without using that fact or a related/derived one? I can of course ...
9
votes
3answers
1k views

Proving that $x(1-e^{-1/x})$ is strictly increasing

Prove that the function below is strictly increasing $$f(x)=x(1-e^{-1/x}), \quad x>0$$
7
votes
4answers
10k views

Proving that $\lim_{h\to 0 } \frac{b^{h}-1}{h} = \ln{b}$

Is there a formal proof of this fact without using L'Hôpital's rule? I was thinking about using a proof of this fact: $$ \left.\frac{d(e^{x})}{dx}\right|_{x=x_0} = e^{x_0}\lim_{h\to 0} \frac{e^{h}-1}{...
12
votes
4answers
544 views

show $\lim_{x\to 0}\frac{e^x-1}{x}=1$ without L'Hopital

how would you show that $$\lim_{x\to 0}\frac{e^x-1}{x}=1$$ without using derivatives or l'hopital but using basic ideas that are generally introduced just before derivatives in a typical introductory ...
4
votes
2answers
375 views

show that if $n\geq1$, $(1+{1\over n})^n<(1+{1\over n+1})^{n+1}$

I have derived the inequality if $k>1$, ${n(n-1)⋯(n-k+1)\over k!} ({1\over n})^k<{(n+1)n⋯(n-k+2)\over k!} ({1\over n+1})^k$ But, my problem is how to use this inequality to prove that if $n\...
4
votes
3answers
234 views

Particular definition of e

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, ...
6
votes
4answers
877 views

proof of derivative of an exponential function

I was told to assume that $$\ln b=\lim_{h\to 0} \frac{\left(b^h-1\right)}{h}$$ where b is a positive, real, base. Unfortunately, being told to assume something isn't good enough. When using L'...
25
votes
1answer
876 views

Elementary proof that monotone functions are differentiable somewhere

It is well-known that every monotone function $f : \mathbb{R} \to \mathbb{R}$ is differentiable almost everywhere (with respect to Lebesgue measure). It is also known if $E$ has measure $0$, then ...
3
votes
4answers
229 views

exponential difference inequality

Im asked to prove the inequality when $0\leq a<b$ and $x>0$: $$ a^x(b-a)<{b^{x+1}-a^{x+1}\over{x+1}}<b^x(b-a) $$ So far I have seen that obviously: $$a^x(b-a)<b^x(b-a)$$ and that $$b^{...

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