Question

(lim x->0) { [(1+x)^1/x - e - ex/2]/ x^2}

Answer 1

\[\lim_{x \rightarrow 0} \frac{ [ (1+x)^\frac{ 1 }{ x } -e +\frac{ ex }{ 2 } ] }{x^2}\]

Answer 2

I tried expanding the series but with no luck. \[(1+x)^\frac{ 1 }{ x } \] Can this series be expanded without taylor's expansion?

Answer 3

well as \(x\to0\) we have \((1+x)^{1/x}\to e\)

Answer 4

right the function tends to 0/0 so L'hospital rule should work here, but the derivative was not clear.

Answer 5

You cannot use l'hopitals here yet.

Answer 6

If what @oldrin.bataku said isn't obvious then once we make the substitution:\[\bf h=\frac{1}{x} \implies \lim_{x \rightarrow 0}(1+x)^{1/x}=\lim_{h \rightarrow \infty }\left( 1+\frac{ 1 }{ h } \right)^h=e\]

Answer 7

Because of this the question can be simplified to:\[\bf \lim_{x \rightarrow 0}\frac{ \frac{ ex }{ 2 } }{ x^2 }=\frac{ e }{ 2 }\lim_{x \rightarrow 0}\frac{ 1 }{ x }\]

Answer 8

lim y=(1+x)^(1/x)
ln(lim y) = (1/x)ln(1+x)
the rhs is 0/0
use lahopital

ln(lim y) = (1/(1+x))
take limit on rhs
ln(lim y) = 1
e^1 = lim y

Answer 9

so after probably an hour of playing around with a multitude of approaches I think I got it playing with Taylor series

Answer 10

good man^^^

Answer 11

does it have a finite limit?

Answer 12

indeed :-) although I cheated by knowing what to expect by using wolfram|alpha though

Answer 13

Okay, prepare for a bunch of playing around with series. Observe that since our limit involves dividing by \(x^2\), I will play with the series and only pay attention to the terms that are \(O(x^2)\) as higher-degree terms will pass through and tend to \(0\) as \(x\to0\) while constants and negative-degree terms risk blowing up.

Answer 14

First, note we have \((1+x)^{1/x}\) in our limit and this is probably the most problematic expression in our limit; we want to get rid of it. Observe, however, that we can rewrite it in terms of its logarithm (just like you do in logarithmic differentiation):$$(1+x)^{1/x}=e^{\log(1+x)^{1/x}}=e^{1/x\log(1+x)}$$aha, note we know of a series for \(\log(1+x)\):$$\log(1+x)=\sum_{k=1}^\infty\frac{(-1)^{k+1}}kx^k\\\frac1x\log(1+x)=\sum_{k=0}^\infty\frac{(-1)^k}{k+1}x^k=1-\frac12x+\frac18x^2+O(x^3)$$

Answer 15

thus we can rewrite \((1+x)^{1/x}\) in terms of this series:$$(1+x)^{1/x}=\exp\left(1-\frac12x+\frac18x^2+O(x^3)\right)=e\exp\left(-\frac12x\right)\exp\left(\frac18x^2\right)\exp(O(x^3))$$

Answer 16

oops the \(\dfrac18 x^2\)s above should all be \(\dfrac13x^2\)

Answer 17

Now, recall that we know a convenient series for \(\exp x\):$$\exp x=\sum_{k=0}^\infty\frac1{k!}x^k=1+x+\frac12x^2+O(x^3)\\\exp\left(-\frac12x\right)=1-\frac12x+\frac18x^2+O(x^3)\\\exp\left(\frac18x^2\right)=1-\frac18x^2+O(x^3)\\\exp(kx^n)=1+O(x^3)\text{ for }n\ge3$$

Answer 18

gah that should be:$$\exp\left(\frac13x^2\right)=1+\frac13x^2+O(x^3)$$ sorry!!

Answer 19

We can now replace those above exponentials with our series:$$(1+x)^{1/x}=e\left(1-\frac12x+\frac18x^2+O(x^3)\right)\left(1+\frac13 x^2+O(x^3)\right)(1+O(x^3))$$
Now observe that when we multiply only a few significant terms will remain:$$ (1+x)^{1/x}=e\left(1-\frac12 x+\frac18x^2+\frac13x^2+O(x^3)\right)=e\left(1-\frac12x+\frac{11}{24}x^2\right)+O(x^3)$$

Answer 20

Now we're ready to substitute our series into our limit:$$\lim_{x\to0}\frac{(1+x)^{1/x}-e+ex/2}{x^2}=\lim_{x\to0}\frac{e\left(1-\frac12x+\frac{11}{24}x^2-1+\frac12x\right)+O(x^3)}{x^2}$$Observe as our limit drastically reduces:$$\lim_{x\to0}\left(\frac{e\left(\frac{11}{24}x^2\right)}{x^2}+O(x)\right)=\lim_{x\to0}\frac{11}{24}e=\frac{11}{24}e$$

Answer 21

The majority of that derivation involved determining a power series for \((1+x)^{1/x}\)... there may be an easier way that I didn't think about :-p