Question

Prove that limit of (1+a^x)^1/x @ x tending to 0 is e^a

Answer 1

\[\lim_{x \rightarrow 0}(1+a^x)^(1/x)=e^a\]

Answer 2

\[\lim_{x \rightarrow 0}(1+a^x)^{1/x}=e^a\]

Answer 3

ya !

Answer 4

not true

Answer 5

you can try using l'hopital and see

Answer 6

It is 100% true

Answer 7

http://www.wolframalpha.com/input/?i=lim+x-%3E0+%281%2Ba^x%29^%281%2Fx%29

Answer 8

How?

Answer 9

really...can't wait to see the proof :)

Answer 10

i think it is zero

Answer 11

no

Answer 12

oops wolfram beat me again. depends on the direction

Answer 13

1/x -> inf
a^x = 1

Answer 14

what have u done

Answer 15

i think maybe the quesion is
\[\lim_{x\to 0}(1+ax)^{\frac{1}{x}}\] then you get \(e^a\)

Answer 16

no

Answer 17

but if u prove that one i will do this

Answer 18

ok sorry i think i have wrong question

Answer 19

imagine

Answer 20

You might want to try finding this limit\[\large \lim_{x \rightarrow \infty} \left(1+{1 \over x}\right)^{ax}\]

Answer 21

ok!will u?

Answer 22

one line
\[e=\large \lim_{x \rightarrow \infty} \left(1+{1 \over x}\right)^{x}\implies e^a=\large \lim_{x \rightarrow \infty} \left(1+{1 \over x}\right)^{ax}\]

Answer 23

I'll get you started at least. First, let\[\large y= \lim_{x \rightarrow \infty} \left(1+{1 \over x}\right)^{ax}\]Now you need to take the natural log of both sides. \[\large \ln(y)= \lim_{x \rightarrow \infty} \ln\left(\left(1+{1 \over x}\right)^{ax}\right)\]Simplify a bit, and then use L'hopitals Rule.

Answer 24

wait please

Answer 25

Or this should also work out
\[ \huge e^{ \lim_{x->0}\ln(1+\frac{1}{x})^{ax}}\]

Answer 26

then let 1/x = h, h-> inf

Answer 27

depends on your definition of \(e\) maybe
but one definition is \(e=\lim_{x\to \infty}(1+\frac{1}{x})^x\) in which case it is immediate

Answer 28

I can't do that king george please tell me how to do it coz i am only beginning the limit topic

Answer 29

\[\large \ln(y)= \lim_{x \rightarrow \infty} \ln\left(\left(1+{1 \over x}\right)^{ax}\right)\]\[\large \ln(y)= \lim_{x \rightarrow \infty} (ax)\cdot\ln\left(1+{1 \over x}\right)\]This is an indeterminate of the form \(\infty\cdot 0\) so we can use L'hopitals rule on both sides.

Answer 30

k! I try

Answer 31

Good luck.

Answer 32

thanx!

Answer 33

@kingGeorge
\[
\large \ln(y)= \ (ax)\cdot\ln\left(1+{1 \over x}\right)=\\
\frac {\ln\left(1+{1 \over x}\right)} {\frac 1{ax}}

\]
Apply L'Hospital's rule again.

Answer 34

why did u divide by 1/ax.

Answer 35

to get the form 0/0 when x tends to infinity

Answer 36

have u taken derivative?

Answer 37

i m getting confused .i don't know anything please solve fully.

Answer 38

Derivative of Numerator is
\[
-\frac{1}{\left(\frac{1}{x}+1\right) x^2}
\]

Drivative of Denominator is
\[
-\frac{1}{a x^2}
\]

It should be easy from now on.

Answer 39

The ration of the derivatives after simplifcation becomes
\[ \frac{a}{\frac{1}{x}+1}

\]

Answer 40

ratio

Answer 41

k!

Answer 42

I got thanx!