Prove that
$$\lim_{x\to 0} \frac{\ln(x+1)}{x} = 1$$
without using L'Hôpital Rule.
Asked
Active
Viewed 317 times
2
-
1That will be hard to do. – David Mitra Apr 10 '14 at 12:55
-
2@DavidMitra especially because it's wrong. The limit is $0$. $ln(x) = o(x)$ – T_O Apr 10 '14 at 13:03
-
I can help you if the question is proving $$\lim_{x\to0} \frac{\ln(x+1)}{x} = 1.$$ – Tunk-Fey Apr 10 '14 at 13:11
-
I suppose that your question contains a typo. Would it be $$\lim_{x\to0} \frac{\ln(x+1)}{x} = 1$$ – Claude Leibovici Apr 10 '14 at 13:14
-
not sure, but the question given to me is infinity and if u use lopital rule it does give 1 – user10024395 Apr 10 '14 at 13:32
-
1Are you sure you know how to use l'hôspital rule ? $1/x$ goes to 0 when $x$ goes to $+\infty$ – T_O Apr 10 '14 at 13:40
-
ok then what about the case when x tends to 0? – user10024395 Apr 10 '14 at 13:50
-
If you mean @Tunk-Fey limit then calculate $\int_{0}^{x}{\frac{1}{(y+1)}}dy$. Now get an upper and lower bound for this integral (think about areas) and go from there. – Paul Apr 10 '14 at 13:54
-
4Which definition of the logarithm are you working with? – Daniel Fischer Apr 10 '14 at 13:56
-
I agree with Daniel Fischer. Your question is clearly dependent on how you define the logarithm (hence, what properties of the logarithm are assumed). That is because your question is a basic property of the logarithm (which, in terms of derivatives, is the derivative of $\log(1+x)$ at $x=0$. – digital-Ink Apr 10 '14 at 14:23
6 Answers
4
This is just the derivative of $y=lnx$ at $x=1$ by using the classic definition of the derivative. And so the answer is $1$. In this way, L'Hospital is avoided.
imranfat
- 10,029
-
-
1Not really, only a whiff of ambiguity since we don't know how the OP defines log. – Mikhail Katz Apr 10 '14 at 14:10
2
If you define $\log(1+x)$ as the integral of $\frac{1}{1+x}$ and you don't want to use the fundamental theorem of calculus either, you could bound $\frac{1}{1+x}$ between $1-\epsilon$ and $1+\epsilon$ for sufficiently small $x$, and use this to get the required limit.
Mikhail Katz
- 42,112
- 3
- 66
- 131
1
HINT :
If that limit you are asking, then you can use Maclaurin series for $\ln(1+x)$ to prove $\lim\limits_{x\to0} \frac{\ln(1+x)}{x} = 1.$
$$\\$$
$$\Large\color{blue}{\text{# }\mathbb{Q.E.D.}\text{ #}}$$
Tunk-Fey
- 24,849
-
1I believe if he can use McLaurin series he would not bother with avoiding l'Hôspital Rule and just go straight for an equivalent – T_O Apr 10 '14 at 14:04
-
-
1
-
@T_O I don't necessarily think he wouldn't know about Taylor series. Some calc books don't mention l'Hospital until doing sequences and series. – Josh Apr 10 '14 at 14:08
-
1Why on earth are you putting a huge "Q.E.D." after something that is a hint not a proof? – hmakholm left over Monica Apr 10 '14 at 17:04
-
@HenningMakholm Sorry if my Q.E.D. sign is bothering you. I put it behind of every my answer, even if only a hint, just as my signature. It's kind of a personal touch. I hope you don't mind. :) – Tunk-Fey Apr 10 '14 at 17:11
1
Integrate $\frac{1}{1+y}$ between 0 and x. Look at the picture below to get an upper and lower bound for this integral. Divide through by x and take the limit.
Paul
- 1,482
1
This limit is equivalent to $$ \lim_{x \to 0} \frac{e^x-1}{x}=1, $$ which is easily handled with the definition $$ e^x = \sum_{k=0}^\infty \frac{x^k}{k!}. $$ As noticed above, it all boils down to your definition of the exponential or the logarithmic function.
Siminore
- 35,136
0
For any $x>0$ We have $$\frac{x}{x+1} =\int_{0}^x\frac{dt}{x+1} \le \int_{0}^x\frac{dt}{t+1} =\color{red}{\ln(x+1 )}=\int_{0}^x\frac{dt}{t+1} \le \int_{0}^x\frac{dt}{1} = x $$
that is for $x>0$
$$\frac{x}{x+1} \le \ln(x+1 ) \le x \implies\color{red}{\frac{1}{x+1}\le \frac{\ln(x+1 )}{x} \le 1}$$ $$\color{red}{1=\lim_{x\to 0}\frac{1}{x+1}\le \lim_{x\to 0}\frac{\ln(x+1 )}{x} \le 1}$$
that $$\color{blue}{ \lim_{x\to 0}\frac{\ln(x+1 )}{x} = 1}$$
By Squeeze Theorem, we have,
Guy Fsone
- 23,903