Question

find \lim_{t\rightarrow 0}(ax^t+by^t)^{\frac{1}{t}} when a+b=1

Answer 1

\[\large \lim_{t\rightarrow 0}(ax^t+by^t)^{\frac{1}{t}}\]

@UnkleRhaukus ?

Answer 2

When a+b = 1... how did you go about using that substitution?

Answer 3

Series expansion?

Answer 4

@Kainui

Answer 5

1,1/2,inf

Answer 6

Looks like you need to use logarithms.

Answer 7

Yeah

Answer 8

\[\large \lim_{t\rightarrow 0}(ax^t+by^t)^{\frac{1}{t}} = e^{\lim \limits_{t\rightarrow 0} \frac{\ln(ax^t+by^t) }{t}}\]

L'hopital next

Answer 9

you should end up with \(x^ay^b\)
give it a try i dont want to spoil your fun by posting the solution :P

Answer 10

This was quite a fun limit, and usually I don't really care for limits, but this was quite a surprising result.

Answer 11

the solution without L'hopitals is quite nice too
il post it later if somebody is interested..

Answer 12

*Raises hand*

Answer 13

I'm curious @ganeshie8 but I will try to figure it out on my own first give me a few minutes and I'll tell you if I give up.

Answer 14

Mhm sec

Answer 15

I give up, what I tried was making this substitution then using the binomial theorem.\[\Large \lim_{n \to \infty} (ax^{1/n}+by^{1/n})^n = x \sum_{k=0}^\infty \left(\begin{matrix}n \\ k\end{matrix}\right) a^n(\frac{x}{y})^{\frac{k}{n}}(\frac{b}{a})^k\]

Answer 16

Why binomial theorem?

Answer 17

I don't know, I just ran out of ideas. Although I think I messed it up because the n terms inside should still be infinite. It's kind of like how this is false: \[\Large \lim_{n \to \infty}(1+\frac{1}{n})^n \ne \prod_{n=1}^\infty (1+\frac{1}{n})\]

Answer 18

@ganeshie8 what's the answer?

Answer 19

hey Kainui you seem to be having the similar idea as mine

Answer 20

we use below standard limit
\[\lim\limits_{n\to 0 } \left(1+nx\right)^{1/n} = e^x\]

Answer 21

I see that was kind of along the lines of what I was thinking.

Answer 22

\[\begin{align}

\lim \limits_{t\to 0}(ax^t+by^t)^{1/t} &= \lim \limits_{t\to 0}\left[(1-b)x^t+by^t\right]^{1/t} \\
&= x\lim \limits_{t\to 0}\left[1-b+b\left(\frac{y}{x}\right)^t\right]^{1/t} \\

\end{align}\]

Answer 23

That's fascinating, nice.

Answer 24

say \(y/x\) = u for the time being and lets work that piece

\[\begin{align}


\lim \limits_{t\to 0}\left[1-b+bu^t\right]^{1/t}
&= \lim \limits_{t\to 0}\left[1-b+bu^t\right]^{1/t} \\
&= \lim \limits_{t\to 0}\left[1+b(u^t\color{Red}{-} 1)\right]^{1/t} \\

\end{align}\]

Answer 25

the idea is to use another limit
\[\lim_{t\to0}\dfrac{f(x)^t-1}{t}=\ln(f(x))\]

this proof is trivial to, we can work it later i hope

Answer 26

\[
\begin{align}


\lim \limits_{t\to 0}\left[1-b+bu^t\right]^{1/t}
&= \lim \limits_{t\to 0}\left[1-b+bu^t\right]^{1/t} \\
&= \lim \limits_{t\to 0}\left[1+b(u^t\color{Red}{-} 1)\right]^{1/t} \\
&= \lim \limits_{t\to 0}\left[1+bt\frac{(u^t\color{Red}{-} 1)}{t}\right]^{1/t} \\
&= \lim \limits_{t\to 0}\left[1+{b\ln u} t \right]^{1/t} \\
&= e^{{b\ln u}} \\
&= u^b \\

\end{align}

\]

Answer 27

*.*

Answer 28

whered u get this from http://prntscr.com/5ehysy

Answer 29

pretty result

Answer 30

That was nice, but where can I find that proof?

Answer 31

that was from a real analysis problem set
one sec the proof is easy+pretty, let me pull it up

Answer 32

I'll post it as new question