Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 :) ):

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

share|improve this question
1  
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

3 Answers 3

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 $f(x)=e^{-x}$.

share|improve this answer
    
(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 $\,a>0\,$:

$$T:=\frac{a\left(1-e^{-a/2^k}-\frac{a}{2^k}e^{-a/2^k}\right)}{\frac{a}{2^k}}$$

and now make the substitution

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

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

share|improve this answer

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

share|improve this answer
1  
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.