Why is $\log_e(x)'=\ln(x)'= 1/x$? how to prove the differentiation rule. ofc that means also $\int(1/x)dx=\ln(x) + c$  I think that the derivative of the $\ln$ of a variable with respect to the variable being 1/x is an axiom to prove that the limit of   $(1 + 1/(ax))^{bx}$ as $x \to \infty$  is $e^{b/a}$  How to prove one without using the other and any theorem following from them ?  Or at least can we prove one of them without the other and the other should follow?
119803
I'm trying to prove that $\frac{\mathrm{d} }{\mathrm{d} x}\ln x = \frac{1}{x}$.  Here's what I've got so far: $$ \begin{align} \frac{\mathrm{d}}{\mathrm{d} x}\ln x &amp;= \lim_{h\to0} \frac{\ln(x + h) - \ln(x)}{h} \\ &amp;= \lim_{h\to0} \frac{\ln(\frac{x + h}{x})}{h} \\ &amp;= \lim_{h\to0} \frac{\ln(1 + \frac{h}{x})}{h} \\ \end{align} $$ To simplify the logarithm: $$ \lim_{h\to0}\left (1 + \frac{h}{x}\right )^{\frac{1}{h}} = e^{\frac{1}{x}} $$ ^This is the line I have trouble with. I can see that it is true by putting numbers in, but I can't prove it. I know that $e^{\frac{1}{x}} = \lim_{h\to0}\left (1 + h \right )^{\frac{1}{xh}}$, but I can't work out how to get from the above line to that. $$ \lim_{h\to0}\left ( \left (1 + \frac{h}{x}\right )^{\frac{1}{h}}\right )^{h} = e^{\frac{h}{x}} $$ Going back to the derivative: $$ \begin{align} \frac{\mathrm{d}}{\mathrm{d} x}\ln x &amp;= \lim_{h\to0} \frac{\ln(e^{\frac{h}{x}})}{h} \\ &amp;= \lim_{h\to0} \frac{\frac{h}{x}\ln(e)}{h} \\ &amp;= \lim_{h\to0} \frac{h}{x} \div h\\ &amp;= \frac{1}{x} \\ \end{align} $$  This proof seems fine, apart from the middle step to get $e^{\frac{1}{x}}$. How could I prove that part?
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 $0$.
100415
$$\int\frac{(\ln x)^{10}}{x}\,dx$$ All I know is that I am supposed to substitute $u=\ln x$. But can someone please explain to me how to find the anti derivative of $(\ln x)^{10}$. I think we are supposed to use integration by parts on integrating $\ln x$, but we haven't been taught that yet.
21812
I ask for the proof of the L'Hôpital rule for the indeterminate form $\frac{\infty}{\infty}$ utilizing the rule for the form $\frac{0}{0}$.   Theorem: Let $f,g:(a,b)\to \mathbb{R}$ be two differentiable functions such as that:  $\forall x\in(a,b)\ \ g(x)\neq 0\text{ and }g^{\prime}(x)\neq 0$ and $\lim_{x\to a^+}f(x)=\lim_{x\to a^+}g(x)=+\infty$ If the limit $$\lim_{x\to a^+}\frac{f^{\prime}(x)}{g^{\prime}(x)}$$ exists and is finite, then  $$\lim_{x\to a^+}\frac{f(x)}{g(x)}=\lim_{x\to a^+}\frac{f^{\prime}(x)}{g^{\prime}(x)}$$  My attempt:  Since $\lim_{x\to a^+}f(x)=+\infty$, $$\exists \delta&gt;0:a&lt;x&lt;a+\delta&lt;b\Rightarrow f(x)&gt;0\Rightarrow f(x)\neq 0$$  Let $F,G:(a,a+\delta)\to \mathbb{R}$, $F(x)=\frac{1}{f(x)}$, $G(x)=\frac{1}{g(x)}$. Then by the hypothesis $\lim_{x\to a^+}F(x)=\lim_{x\to a^+}G(x)=0$, $$\forall x\in(a,b)\ \ G(x)\neq 0\text{ and }G^{\prime}(x)=-\frac{1}{g^2(x)}g^{\prime}(x)\neq 0$$ The question is, does the limit $$\lim_{x\to a^+}\frac{F^{\prime}(x)}{G^{\prime}(x)}=\lim_{x\to a^+}\frac{-\frac{1}{f^2(x)}f^{\prime}(x)}{-\frac{1}{g^2(x)}g^{\prime}(X)}=\lim_{x\to a^+}\frac{g^2(x)f^{\prime}(x)}{f^2(x)g^{\prime}(x)}$$ exist?   The limit $$\lim_{x\to a^+}\frac{f^{\prime}(x)}{g^{\prime}(x)}$$ exists by the hypothesis but we don't know if the limit $\displaystyle\lim_{x\to a^+}\frac{g^2(x)}{f^2(x)}$ exists to deduce that the limit $$\lim_{x\to a^+}\frac{F^{\prime}(x)}{G^{\prime}(x)}$$ exists to use the L'Hôpital Rule for the form $\frac{0}{0}$.  EDIT: After discussing it with other users in the site, we came to the conclusion that this proof is only partial and can't logically be continued to yield the Theorem.  As a result, the rule for the $\frac{0}{0}$ form can't be used to proove the rule for the  $\frac{\infty}{\infty}$ form. Mr. Tavares and myself have already given two different proofs (with the pretty much the same main idea) of the Theorem in question using Cauchy's Mean Value Theorem. You can read them below. You can also read the proof Rudin gives for a stronger version of the Theorem (that does not suppose that $\lim_{x\to a^+}f(x)=+\infty$) in his book Principle of Meathematical Analysis. If you have any objections in either proofs please let me know. Thank you.
117031
$$\lim\limits_{x\to\infty} \ \frac{\ln \ x}{x}  = 0$$  How can I know this without looking at the graph? What's the easiest way to equate that expression to 0?  I understand that an integer over x approaches 0 as x approaches 0, but since ln x is an increasing function as x approaches infinity, how do we know which effect outweighs the other?
67772
I'm given that $f \in L^1$ and $f(x) \neq 0$ a.e. and trying to prove that for q > 0  $$\lim_{q \to 0} (\int_0^1 |f|^q)^{1/q} = e^{\int_0^1 \ln|f|}$$  My work:  I took the log of both sides to reduce the problem to  $$\lim_{q \to 0} \frac{1}{q} \ln(\int_0^1 |f|^q) = \int_0^1 \ln|f|$$  Since $\ln$ is a concave function I used Jensen's inequality to show that  $$\frac{1}{q} \ln(\int_0^1 |f|^q) \geqslant \int_0^1 \frac{1}{q}\ln|f|^q = \int_0^1 \ln|f| $$  So if the limit exists, I know  $$\lim_{q \to 0} \frac{1}{q} \ln(\int_0^1 |f|^q) \geqslant \int_0^1 \ln|f|$$  Now I can't make any more progress to reverse the inequality.  Maybe I need a completely different approach.  Also what about the necessity of $f \in L^q$? Please help.
65798
$$f(x)=\begin{cases} 0 &amp;:&amp; x=0\\[1ex]x^{\frac{2}{3}} \log x&amp;:&amp;\text{otherwise}\end{cases}$$  My thoughts: $\log x$ behaves weird in $(0;1)$ and will not be bounded for any $ x \in V_{\delta} (0) , \delta &gt;0 $  Now is $ x^{\frac{2}{3}}$ small enough to keep $\log x$ in control? How can I show this? Please help. I'm looking for a proof using the definition of continuity.
24630
I need to prove the  continuity of $f(x)=\log x$ using a $\epsilon-\delta$ proof  These is what I have so far but am not sure how to continue  $|\log x-\log a| &lt; \epsilon$   $\log a- \epsilon &lt; \log x &lt; \log a+ \epsilon$  $\frac{a}{e^\epsilon} &lt; x &lt; {a}e^\epsilon$  Any help is appreciated
8668
Show that $$\lim_{x \to \infty} \frac{\log(1 + e^x)}{x} = 1$$ How do I prove this? Or how do we get this result? Here $\log$ is the natural logarithm.
54356
What is the proof for the power rule for logarithms? And are there different ways to prove it?
85831
By L'Hôpital's rule, it gets $$\lim_{x\to 0}-\frac{\left(e^{(x+1)^{1/x}} (x+1)^{1/x}-e (x+1)^{e/x}\right) ((x+1) \log (x+1)-x)}{2x^3(1+x)},$$which becomes more complicated.  What could I do then?
92698
I conjecture that in some specific conditions a differentiating function gives the following equality: $$\lim_{n\to\infty} f(n+1) - f(n) = \lim_{x\to\infty} (f(x))' $$  However, I'm not sure yet what exactly those conditions are in order to precisely know where I may apply this rule or not. If you wanna take a look over my posted problem  you'll immediately notice that this rule applies for that case. I really appreciate if you help me clarify this.
60255
My brother brought me this same question, also on this .   $$\int_{0}^{x}f(t)dt = x+\int_{x}^{1}tf(t)dt \tag{1}$$  When I try to solve it, I also did same thing. Differentiation leads to  $$f(x) = 1-xf(x)\\  f(x) = \frac{1}{1+x} \\ \int_{0}^{1}f(x) dx = \ln 2 $$  But try putting $x = 1$ in original equation $(1)$,  $$\int_{0}^{1} f(t) dt = 1$$  so what has happened? We get two contradictory results.  Is it still true that $f(1) = 1/2$? Where is the mistake?
27738
Prove that      $$ \int_{0}^{\infty}\frac{e^{-bx}-e^{-ax}}{x}\,dx = \ln\left(\frac{a}{b}\right) $$   My Attempt:  Define the function $I(a,b)$ as  $$ I(a,b) = \int_{0}^{\infty}\frac{e^{-bx}-e^{-ax}}{x}\,dx $$  Differentiate both side with respect to $a$ to get  $$ \begin{align} \frac{dI(a,b)}{da} &amp;= \int_{0}^{\infty}\frac{0-e^{-ax}(-x)}{x}\,dx\\ &amp;= \int_{0}^{\infty}e^{-ax}\,dx\\ &amp;= -\frac{1}{a}(0-1)\\ &amp;= \frac{1}{a} \end{align} $$  How can I complete the proof from here?
106621
I am trying to calculate the following limit without L'Hôpital's rule:     $$\lim_{n \to \infty} \dfrac{n}{\ln(n)}$$   I tried every trick I know but nothing works. You don't have to prove it by definition.
112982
Reasoning for $\frac{\ln|x|}{2}+ C$ :  $$\int  \frac{1}{(2x)}  \, dx = \frac{\ln|2x|}{2}+ C$$   Check: By taking derivative we get $\frac{1}{x}$  Reasoning for $\frac{\ln|x|}{2}+ C$ :  $$\int  \frac{1}{2x}  \, dx = \frac{1}{2}\int  \frac{1}{x}  \, dx =\frac{\ln|x|}{2}+ C$$  Check: By taking derivative we get $\frac{1}{x}$  Question: Is  $\int  \frac{1}{2x}  \, dx $ equal to $\frac{\ln|2x|}{2}+ C$ or $\frac{\ln|x|}{2}+ C$?
109742
I have been trying to prove this by induction on $n\in \mathbb{N}$, but this approach seemed to get me nowhere. I have a suspicion it might be necessary to express $\log{n}$ as $\int_1^n 1/x\text{ }dx$, but I could not build a rigorous argument from that.  Any help greatly appreciated!