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$$\lim_{x\to 0} \frac{\ln(1+x)}x$$

The process I want to take to solving this is by using the definition of the limit, but I am getting confused. ( without l'hopitals rule)

$$\lim_{h \to 0} \frac{f(x+h) - f(x)}h$$

$$\lim_{h \to 0} \frac{\frac{\ln (1+x+h)}{x+h} - \frac{\ln(1+x)}x}h$$

$$\lim_{h \to 0} \frac{x\ln(1+x+h) - (x+h)\ln (1+x)}{hx(x+h))}$$

At this point I get confused because I know the answer is $1$, but I am not getting this answer through simplification of my formula.

rubik
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heyjude
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  • Can you use that $(1+x)^{1/x}\to e$ as $x\to 0$? – Pedro Feb 26 '14 at 07:11
  • I think you may be mixing up the definition of the limit and the definition of differentiation (note that $\lim_{h \to 0} \frac{f(x+h)-f(x)}{h}$ refers to the differential at $x$. Instead the first question is about the limit at $0$.) If we want to avoid L'hospital rule and go straight to the definition we might want to use a $\delta$-$\epsilon$ argument. – Kelvin Soh Feb 26 '14 at 07:25
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    You would treat "x" as "h", with the function $ \ f(x) = \ln x \ $ , taking $ \ a = 1 \ . $ So you have $ \ [\ln x]' \vert_{x=a=1} = \lim_{h \rightarrow 0} \ \frac{\ln(1+h) - \ln 1}{h} \ . $ – colormegone Feb 26 '14 at 07:30
  • The evaluation of this limit depends on the definition of $\ln x$ and in case you use the definition $\ln x = \int_{1}^{x}(dt/t)$ then the limit represents derivative of $\ln x$ at $x = 1$ and therefore is equal to $1$. – Paramanand Singh Feb 26 '14 at 10:49

3 Answers3

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Introduce a new variable $u = 1/x$. Then you limit becomes

$$\lim_{u\to\infty} \frac{\ln(1 + 1/u)}{1/u}$$

mulitply numerator and denominator by $u$, you get

$$\lim_{u\to\infty} u [ \ln(1 + 1/u) ] $$

move $u$ into the $\log$, getting

$$\lim_{u\to\infty} [ \ln(1 + 1/u)^u ]$$

then, since $\ln$ is continuous

$$\ln \left( \lim_{u\to\infty} (1 + 1/u)^u \right)$$

limit inside is equal to $e$ and $\ln(e) = 1$

Winther
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Piotr
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9

You are talking about L'Hôpital's rule, so I assume you already know how to differentiate the logarithm. Now note, that

$$\frac{\log(x+1)}x = \frac{\log(x+1)-\log(1)}{(x+1)-1}$$

Thus

$$\lim_{x\to0}\frac{\log(x+1)}x = \lim_{x\to0}\frac{\log(x+1)-\log(1)}{(x+1)-1}=\left(\log(x)\right)^\prime_{x=1}=\left.\frac{1}x\right|_{x=1}=1$$

(This is not by using L'Hôpital's rule but only by using the definition of derivative and knowing the derivative of $\log(x)$)

user127.0.0.1
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    The derivative of $\ln x$ is often derived by using the limit in question, so this argument may or may not be circular (depending on how one has introduced the logarithm function). I don't think this question is really well-posed unless the OP tells us exactly what properties of the logarithm that we are allowed to use... – Hans Lundmark Feb 26 '14 at 07:40
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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

$$\bbox[5px,border:2px solid #C0A000]{\frac{x-1}{x}\le \log(x)\le x-1} \tag 1$$

for $x>0$.

Hence, we have from $(1)$

$$\frac{1}{1+x}\le \frac{\log(1+x)}{x}\le 1$$

whereupon application of the squeeze theorem yields the coveted result

$$\lim_{x\to 0}\frac{\log(1+x)}{x}=1$$

Community
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Mark Viola
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