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Let $\lambda>0$ and look at:

$$\lim _{k \to \infty}\frac{\lambda \cdot (1-e^{-\lambda/2^k}-\frac{\lambda}{2^k}e^{-\lambda/2^k})}{\frac{\lambda}{2^k}}$$

I know it's zero (long live wolfram alpha), but I really can't see why. Can someone please help me.

Or maybe equivalently:

$$\lim _{h \to 0}\frac{1-e^{h}-he^h}{h}$$

Oh sorry it missed two minuses to be correct, it should have been (but I can figure out that one now :) ):

$$\lim _{h \to 0}\frac{1-e^{-h}-he^{-h}}{h} $$

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I don't think your original limit is equivalent to what you wrote: you didn't write the minus signs in the exponents! – DonAntonio Jan 7 '13 at 12:06
up vote 3 down vote accepted

Edited to fit the question.

Use that $$ \lim_{h\to 0}\frac{1-e^{-h}}{h}=-\lim_{h\to 0}\frac{e^{-h}-e^{-0}}{h-0}=-f'(0)=1, $$ where $f(x)=e^{-x}$.

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(Internal:) Haha, når du ikke er på dit kontor længere må jeg finde dig på nettet :D – Henrik Jan 7 '13 at 12:02
Haha, alletiders ;) – Stefan Hansen Jan 7 '13 at 12:05

Instead of lambda let us use $\,a>0\,$:


and now make the substitution

$$x:=\frac{a}{2^k}\,\,\,,\,,\text{ so that}\,\,\, k\to\infty\Longrightarrow x\to 0\,\,\longrightarrow$$

$$\lim_{k\to\infty} T=\lim_{x\to 0}\frac{a\left(1-e^{-x}-xe^{-x}\right)}{x}\stackrel{\text{L'Hospital}}=\lim_{x\to 0}a\left(e^{-x}-e^{-x}+xe^{-x}\right)=\lim_{x\to 0}axe^{-x}=0$$

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$$\lim _{h \to 0}\frac{1-e^{h}-he^h}{h}=$$ (L'Hopital rule) $$=\lim _{h \to 0}\frac{-e^{h}-e^h-he^h}{1}=\lim_{h\to0}-2e^h-he^h=-2$$ after you changed question to

$$\lim _{h \to 0}\frac{1-e^{-h}-he^{-h}}{h}=$$ (L'Hopital rule) $$=\lim _{h \to 0}\frac{e^{-h}-e^{-h}+he^{-h}}{1}=\lim_{h\to0}he^{-h}=0$$

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In your last line you must chance the exponents sign... – DonAntonio Jan 7 '13 at 12:20
Thank you Don. it is done – Adi Dani Jan 7 '13 at 12:26

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