Calculate $\lim\limits_{y\to{b}}\frac{y-b}{\ln{y}-\ln{b}}$

How can we locate $\displaystyle \lim_{y\to{b}}\frac{y-b}{\ln{y}-\ln{b}}$ without making use of :

(a) L'Hôpital is regulation, (b) the restriction $\displaystyle \lim_{h \to 0}\frac{e^h-1}{h} = 1$, and also (c) the reality that $\displaystyle \frac{d}{dx}\left(e^x\right) = e^x$.

The factor for the problems is that with this restriction I'm attempting to confirm (c), and also I've done so with (b) and also I collect it would certainly be round to make use of (a). To make sure that is that. Additionally, I would certainly value if you can share several means of confirming that the by-product of $e^x$ is $e^x$. Many thanks a whole lot for your time.

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2019-12-02 02:50:16
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Answers: 4

Say $y = b(1+\epsilon)$. After that $\ln y - \ln b \approx \epsilon$ whereas $y - b = b\epsilon$. So all of it come down to revealing $\ln (1+\epsilon) \approx \epsilon$.

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2019-12-03 04:22:16
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If $e^x=\sum_{n\geq0}x^n/n!$, after that you can show conveniently that $e^{x+y}=e^xe^y$. After that $$\frac{e^{x+h}-e^x}{h} = \frac{e^h-1}{h}e^x,$$ and also in conclusion that $\frac{d}{dx}e^x=e^x$ we require just after that show that $$\frac{e^h-1}{h}\to1$$ if $h\to0$. Utilizing your definition, that is very easy.

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2019-12-03 04:21:45
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Let us consider the restriction in the title and also the definition of $\log x$ (please forgive me for making use of $\log$ symbols as opposed to $\ln$) as in your comment, $$\log x =\int_1^x \frac{1}{t}dt$$ By the basic theory of calculus we have $$\frac{d}{dx}\log x = \frac{1}{x}$$ on the various other hand $$f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}=\lim_{y\to x}\frac{f(y)-f(x)}{y-x}$$ therefore $$\lim_{y\to x}\frac{y-x}{\log y-\log x}=\lim_{y\to x}\frac{1}{\frac{\log y-\log x}{y-x}}=\frac{1}{1/x}=x$$ where in the last action we made use of the ratio regulation $$\lim_{y\to a}\,g(y)=A\, \text{ and }\,\lim_{y\to a}\,h(y)=B\ne0\text{ implies }\lim_{y\to a}\,\frac{g(y)}{h(y)}=\frac{A}{B}$$ with $g(y)=1$.

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2019-12-03 04:21:26
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We can transform your $\lim $ right into the inverse of the by-product of $f(y)=\ln y $, reviewed at $y=b$

$$\underset{y\rightarrow b}{\lim }\dfrac{y-b}{\ln y-\ln b}=\dfrac{1}{\underset{ y\rightarrow b}{\lim }\dfrac{\ln y-\ln b}{y-b}}=\dfrac{1}{\dfrac{d}{dy}\left. \ln y\right\vert _{y=b}}=\dfrac{1}{\left. \dfrac{1}{y}\right\vert _{y=b}}=% \dfrac{1}{\dfrac{1}{b}}=b.$$

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2019-12-03 02:33:26
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