I have $\lim\limits_{x \to 3} \frac{3-x}{\ln(4-x)}=1$
For $x \to 3$, I get: $\frac{0}{0}$
How to calculate it, without L'Hôpital's rule?
Why $\lim\limits_{x \to 3} \frac{3-x}{\ln(4-x)}=1$?
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limits-without-lhopital
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0Have you heard about approximation with Taylor polynomials? – 2017-01-09
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0I try to divide by $x$. – 2017-01-09
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0It is one over the average value of $1/t$ as $t$ ranges from $1$ to $4-x$, while $4-x$ approaches $1$, because $\ln(x)=\int_1^x \frac1t\,dt$. – 2017-01-09
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0let $\ln(1+t)
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0@divisor Tell us how much you know! – 2017-01-09
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0To allow us to produce an acceptable solution, you must provide a definition of the function $\ln(x)$. – 2017-01-09
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0This is a pretty pathetic question; it lacks any show of effort, reference to context, etc., yet it received three upvotes, from some among the crowd that raced to answer it, just because it is among the few questions they can answer, and they care nothing of the quality of a question, and certainly not the quality of the site. Sad, sad, sad...and getting worse. – 2017-01-10
4 Answers
1
$$\lim\limits_{x \to 3} \frac{3-x}{\ln(4-x)}$$
Let $t = 3 - x$
$$\lim\limits_{t \to 0} \frac{t}{\ln(1+ t)} = \lim\limits_{t \to 0} \frac{t/t}{\ln(1+ t)/t} = 1$$
Proof :-
$\displaystyle \ln(x + 1)/x = y \implies 1+x = e^{xy} \implies {(e^{xy} - 1)y\over xy} = 1 \implies \lim_{xy \to 0 } {(e^{xy} - 1)y\over x} \times \lim_{x \to 0}y = 1 \implies \lim_{x \to 0}y = 1$
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0Sorry, What have I missed ? – 2017-01-09
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1I think you have to prove that the limit of $\ln(1+x)/x$ is actually one – 2017-01-09
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0Great thanks :) – 2017-01-09
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0@nbubis Isn't it a standard limit that everyone learns in high school. – 2017-01-09
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1I don't know, but proving that without L'Hôpital's rule is kinda the question here. – 2017-01-09
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0@nbubis check my proof, $(e^x - 1)/x, x\to 0 = 1$ can be proven by sandwich. If this still does not satisfy, Then I am open to delete my answer because beyond this my knowledge is limited. – 2017-01-09
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0Yes, I know $\frac{\ln(1+x)}{x}=1$ – 2017-01-09
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1You are trading the limit of $\ln(1+x)/x$ for that of $(e^x-1)/x$. This leaves a taste of an incomplete argument (just as my limit of $(1+1/n)^n$). Actually, nothing can be conclusive as long as the OP hasn't given any definition. – 2017-01-09
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0$\ln(1+x)/x$ this a bad? – 2017-01-09
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0@divisor Not bad but a bit circular. – 2017-01-09
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0Circular? Wow, why? – 2017-01-09
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0This a logarithm $(1+x)^{\frac1x}=1$, for $x \to 0$, or not? – 2017-01-09
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0$(1+x)^{\frac1x}=1$ - is that not proven 101%? :D – 2017-01-09
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0@divisor Everything we used here is completely proven. – 2017-01-09
6
With $3-x=1/n$,
$$\lim\limits_{x \to 3} \frac{3-x}{\ln(4-x)}=\lim_{n\to\infty}\frac1{n\ln\left(1+\dfrac1n\right)}=\lim_{n\to\infty}\frac1{\ln\left(\left(1+\dfrac1n\right)^n\right)}.$$
You should be able to conclude.
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1This is beautiful! – 2017-01-09
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0This will be another problem $(1+\frac1n)^n$ – 2017-01-09
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1@MyGlasses: that depends on what is taken for granted. (This is rarely stated by posters.) – 2017-01-09
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0Why problem? $(1+\frac1n)^n$=1? – 2017-01-09
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0$\frac1n=0$ and $(1+0)^{1}$ – 2017-01-09
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0@divisor $(1+\frac1n)^n, n\to \infty = e$ and $\ln e - 1$ – 2017-01-09
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0Why $e$, may be $1$? – 2017-01-09
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0$\frac1n=0$, yes? For $n \to \infty$ – 2017-01-09
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0And $(1+0)^{\infty}=1$? – 2017-01-09
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0@divisor: $1.00001^{100000}=2.718268237\cdots$. What can that tell you ? – 2017-01-09
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0@divisor $$(1 + 1/n)^n = \sum^n_{k = 0}{n \choose k} 1^{n-k}(1/n)^k$$ and $\sum^n_{k = 0} {n \choose k} 1^{n-k}(1/n)^k, n \to 0 =1/0! + 1/1! + 1/2! + 1/3! \cdots = e$ – 2017-01-09
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0Yes, you right... – 2017-01-09
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0But, if $x \to 3$, why $n \to \infty$? – 2017-01-09
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0@divisor Please accept deepSea's answer or Yves's answer. – 2017-01-09
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0@divisor $n = 1/(x -3)$, $x \to 3\implies x- 3 \to 0 \implies n \to \infty$ – 2017-01-09
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0@A---B, Dear, your answer is good too, I choose between your and Yves's answer... – 2017-01-09
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0Hard to understand :( – 2017-01-09
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0Let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/51464/discussion-between-a-b-and-divisor). – 2017-01-09
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0@YvesDaoust Somebody has downvoted all the answers here :)) – 2017-01-09
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0@A---B Who is it? – 2017-01-09
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0@divisor You can never tell who has downvoted. That's why I rarely downvote. – 2017-01-09
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0I'm very sorry, your answers really helped me... – 2017-01-09
5
Let $y = 3-x$, and use the well-known property of limit about $\log$ that:
$\displaystyle \lim_{ y \to 0} \dfrac{\log(y+1)}{y} = 1$ to get the answer. This can be proven by the squeeze theorem. Note that if $x > 0$, then $\dfrac{x}{1+x} < \ln(x+1) < x$. From this we deduce the limit. We can consider the case $x < 0$ similarly.
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1and how do you prove this? without L'Hospital? – 2017-01-09
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0Do a Maclaurin expansion of log would do the trick ! – 2017-01-09
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0this always needs derivatives, like L'Hospital – 2017-01-09
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0@Dr.SonnhardGraubner No, it does not require use of derivatives or integrals. See my posted solution for a pre-calculus way forward. -Mark – 2017-01-09
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0I'm reading it now :) – 2017-01-09
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In THIS ANSWER, I showed using only the limit definition of the exponential function and Bernoulli's Inequality that the logarithm function satisfies the inequalities
$$\frac{x-1}{x}\le \log(x)\le x-1 \tag 1$$
for $x>0$.
Using $(1)$ we have
$$1=\frac{3-x}{3-x}\le \frac{3-x}{\log(4-x)}\le \frac{3-x}{\frac{3-x}{4-x}}=4-x \tag 2$$
whereupon applying the squeeze theorem yields the coveted limit
$$\lim_{x\to 3}\frac{3-x}{\log(4-x)}=1$$
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0@divisor Here is an efficient way forward that uses pre-calculus only. It does rely on a set of inequalities that I developed in a reference post. And that set of inequalities relies only on the limit definition of the exponential function and Bernoulli's Inequality. Please let me know how I can improve my answer. I really want to give you the best answer I can. -Mark. And Happy New Year! – 2017-01-09
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0Mark, I upvote, and go to read this: http://math.stackexchange.com/questions/1589429/how-to-prove-that-logxx-when-x1/1590263#1590263 – 2017-01-09