Difference between revisions of "Aufgaben:Problem 8"
From Ferienserie MMP2
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''Concerning part a)'' | ''Concerning part a)'' | ||
| − | It is kind of unclear to me how \( \frac{d}{dx} ln(x) \vert_{x=1} = -x\dfrac{(\ln(1)-\ln(1-\frac{x}{k}))}{\frac{x}{k}} \). Wouldn't it be much easier to just: | + | First thing: It is kind of unclear to me how \( \frac{d}{dx} ln(x) \vert_{x=1} = -x\dfrac{(\ln(1)-\ln(1-\frac{x}{k}))}{\frac{x}{k}} \). Wouldn't it be much easier to just: |
$$ \lim_{k \rightarrow \infty} \ln((1-\frac{x}{k})^{k}) = \lim_{k \rightarrow \infty} k\ln(1-\frac{x}{k}) = \lim_{k \rightarrow \infty} \frac{\ln(1-\frac{x}{k})}{\frac{1}{k}} $$ | $$ \lim_{k \rightarrow \infty} \ln((1-\frac{x}{k})^{k}) = \lim_{k \rightarrow \infty} k\ln(1-\frac{x}{k}) = \lim_{k \rightarrow \infty} \frac{\ln(1-\frac{x}{k})}{\frac{1}{k}} $$ | ||
Revision as of 22:53, 28 December 2014
Concerning part a)
First thing: It is kind of unclear to me how \( \frac{d}{dx} ln(x) \vert_{x=1} = -x\dfrac{(\ln(1)-\ln(1-\frac{x}{k}))}{\frac{x}{k}} \). Wouldn't it be much easier to just:
$$ \lim_{k \rightarrow \infty} \ln((1-\frac{x}{k})^{k}) = \lim_{k \rightarrow \infty} k\ln(1-\frac{x}{k}) = \lim_{k \rightarrow \infty} \frac{\ln(1-\frac{x}{k})}{\frac{1}{k}} $$
and then use Bernoulli-L'Hopital to get:
$$ \lim_{k \rightarrow \infty} \frac{\ln(1-\frac{x}{k})}{\frac{1}{k}} = \lim_{k \rightarrow \infty} \frac{\frac{x}{k^2}}{(\frac{x}{k} - 1) \frac{1}{k^2}} = -x $$
Second thing: Nevermind, confused myself...
Greetings A.
