Why $\lim\limits_{x \to 3} \frac{3-x}{\ln(4-x)}=1$?

Question

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?

It is one over the average value of $1/t$ as $t$ ranges from $1$ to $4-x$, while $4-x$ approaches $1$, because $\ln(x)=\int_1^x \frac1t,dt$. — Jonas Meyer, Jan 09 '17 at 20:12
To allow us to produce an acceptable solution, you must provide a definition of the function $\ln(x)$. — , Jan 09 '17 at 20:28
This 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. — amWhy, Jan 10 '17 at 18:20

score 6 · Answer 1 · answered Jan 09 '17 at 20:18

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.

answered Jan 09 '17 at 20:18

1

This is beautiful! – YukiJ Jan 09 '17 at 20:22
This will be another problem $(1+\frac1n)^n$ – Nosrati Jan 09 '17 at 20:23
1

@MyGlasses: that depends on what is taken for granted. (This is rarely stated by posters.) – Jan 09 '17 at 20:25
Why problem? $(1+\frac1n)^n$=1? – Dave Jan 09 '17 at 20:25
$\frac1n=0$ and $(1+0)^{1}$ – Dave Jan 09 '17 at 20:26
@divisor $(1+\frac1n)^n, n\to \infty = e$ and $\ln e - 1$ – Jan 09 '17 at 20:27
Why $e$, may be $1$? – Dave Jan 09 '17 at 20:28
$\frac1n=0$, yes? For $n \to \infty$ – Dave Jan 09 '17 at 20:29
And $(1+0)^{\infty}=1$? – Dave Jan 09 '17 at 20:30
@divisor: $1.00001^{100000}=2.718268237\cdots$. What can that tell you ? – Jan 09 '17 at 20:32
@divisor $$(1 + 1/n)^n = \sum^n_{k = 0}{n \choose k} 1^{n-k}(1/n)^k$$
and $\sum^n_{k = 0} {n \choose k} 1^{n-k}(1/n)^k, n \to 0 =1/0! + 1/1! + 1/2! + 1/3! \cdots = e$
– Jan 09 '17 at 20:33
Yes, you right... – Dave Jan 09 '17 at 20:35
But, if $x \to 3$, why $n \to \infty$? – Dave Jan 09 '17 at 20:39
@divisor Please accept deepSea's answer or Yves's answer. – Jan 09 '17 at 20:39
@divisor $n = 1/(x -3)$, $x \to 3\implies x- 3 \to 0 \implies n \to \infty$ – Jan 09 '17 at 20:40
@A---B, Dear, your answer is good too, I choose between your and Yves's answer... – Dave Jan 09 '17 at 20:40
Hard to understand :( – Dave Jan 09 '17 at 20:42
Let us continue this discussion in chat. – Jan 09 '17 at 20:42
@YvesDaoust Somebody has downvoted all the answers here :)) – Jan 09 '17 at 20:47
@A---B Who is it? – Dave Jan 09 '17 at 20:50
@divisor You can never tell who has downvoted. That's why I rarely downvote. – Jan 09 '17 at 20:50
I'm very sorry, your answers really helped me... – Dave Jan 09 '17 at 20:53

DeepSea · Answer 2 · 2017-01-09T20:17:13.613

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.

edited Jan 09 '17 at 20:17

answered Jan 09 '17 at 20:12

DeepSea

77,651

1

and how do you prove this? without L'Hospital? – Dr. Sonnhard Graubner Jan 09 '17 at 20:12
Do a Maclaurin expansion of log would do the trick ! – DeepSea Jan 09 '17 at 20:13
this always needs derivatives, like L'Hospital – Dr. Sonnhard Graubner Jan 09 '17 at 20:14
@Dr.SonnhardGraubner No, it does not require use of derivatives or integrals. See my posted solution for a pre-calculus way forward. -Mark – Mark Viola Jan 09 '17 at 20:43
I'm reading it now :) – Dave Jan 09 '17 at 21:04

score 3 · Answer 3 · edited Apr 13 '17 at 12:21

3

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

$$\lim_{x\to 3}\frac{3-x}{\log(4-x)}=1$$

edited Apr 13 '17 at 12:21

Community

1

answered Jan 09 '17 at 20:42

Mark Viola

179,405

@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! – Mark Viola Jan 09 '17 at 20:46
Mark, I upvote, and go to read this: http://math.stackexchange.com/questions/1589429/how-to-prove-that-logxx-when-x1/1590263#1590263 – Dave Jan 09 '17 at 21:01

score 1 · Accepted Answer · 2017-01-09T20:24:05.737

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 :-

$\displaystyle \ln(x + 1)/x = y \implies 1+x = e^{xy} \implies {(e^{xy} - 1)y\over xy} = 1 \implies \lim_{xy \to 0 } {(e^{xy} - 1)y\over x} \times \lim_{x \to 0}y = 1 \implies \lim_{x \to 0}y = 1$

edited Jan 09 '17 at 20:24

answered Jan 09 '17 at 20:13

Sorry, What have I missed ? – Jan 09 '17 at 20:13
1

I think you have to prove that the limit of $\ln(1+x)/x$ is actually one – Nathaniel Bubis Jan 09 '17 at 20:16
Great thanks :) – Dave Jan 09 '17 at 20:16
@nbubis Isn't it a standard limit that everyone learns in high school. – Jan 09 '17 at 20:19
1

I don't know, but proving that without L'Hôpital's rule is kinda the question here. – Nathaniel Bubis Jan 09 '17 at 20:20
@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. – Jan 09 '17 at 20:26
Yes, I know $\frac{\ln(1+x)}{x}=1$ – Dave Jan 09 '17 at 20:48
1

You 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. – Jan 09 '17 at 21:08
$\ln(1+x)/x$ this a bad? – Dave Jan 09 '17 at 21:26
@divisor Not bad but a bit circular. – Jan 09 '17 at 21:44
Circular? Wow, why? – Dave Jan 09 '17 at 21:59
This a logarithm $(1+x)^{\frac1x}=1$, for $x \to 0$, or not? – Dave Jan 09 '17 at 22:00
$(1+x)^{\frac1x}=1$ - is that not proven 101%? :D – Dave Jan 09 '17 at 22:03
@divisor Everything we used here is completely proven. – Jan 09 '17 at 22:04

Why $\lim\limits_{x \to 3} \frac{3-x}{\ln(4-x)}=1$?

4 Answers4