How do I prove that both are equivalent limits

[sooyong94]

Homework Statement

If k is a positive integer, then show that
$\lim_{x\to\infty} (1+\frac{k}{x})^x = \lim_{x\to 0} (1+kx)^\frac{1}{x}$

Homework Equations

l'hospital's rule, Taylor's expansion

The Attempt at a Solution

How should I begin? Should I prove that both has the same limit, or is there another way to work this question out?

 

[QuantumQuest]

You can do the Taylor expansions to some number of terms for both limits and see - easily enough, that they converge to the same limit.

 

[Mark44]

Homework Statement

If k is a positive integer, then show that
$\lim_{x\to\infty} (1+\frac{k}{x})^x = \lim_{x\to 0} (1+kx)^\frac{1}{x}$

Homework Equations

l'hospital's rule, Taylor's expansion

The Attempt at a Solution

How should I begin? Should I prove that both has the same limit, or is there another way to work this question out?

A simpler approach than the one suggested by QuantumQuest is use a simple substitution.

 

[sooyong94]

What substitution do you mean?

 

[blue_leaf77]

What substitution do you mean?

When you compare
$$(1+\frac{k}{x})^x$$
and
$$(1+kx)^\frac{1}{x},$$
how can you go from one to the other?

 

[Mark44]

What substitution do you mean?

When you compare
$$(1+\frac{k}{x})^x$$
and
$$(1+kx)^\frac{1}{x},$$
how can you go from one to the other?

Maybe think about them like this:
$(1+\frac{k}{x})^x$
and
$(1+ku)^\frac{1}{u}$

 

[sooyong94]

Will this work as a proof?

$L_1=\lim_{x\to\infty} (1+\frac{k}{x})^x$
$\ln L_1=\ln[\lim_{x\to\infty} (1+\frac{1}{k})^x]$
$\ln L_1=\lim_{x\to\infty} \ln[(1+\frac{1}{k})^x]$$\ln(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-...$
$\ln(1+\frac{k}{x})=\frac{k}{x}-\frac{k^2}{2x^2}+\frac{k^3}{3x^3}-...$

$=\lim_{x\to\infty} x [\frac{k}{x}-\frac{k^2}{2x^2}+\frac{k^3}{3x^3}-...]$
$=\lim_{x\to\infty} [k-\frac{k}{2x^2}+\frac{k^2}{3x^3}-...]$
$=k$

Since
$\ln L_1=k$

Therefore
$L_1=e^k$

For the right hand side
$L_2=\lim_{x\to 0} (1+kx)^\frac{1}{x}$
$\ln L_2=\ln\lim_{x\to 0} (1+kx)^\frac{1}{x}$
$\ln L_2=\lim_{x\to 0} \ln (1+kx)^\frac{1}{x}$
$\ln L_2=\lim_{x\to 0} \frac{1}{x} \ln (1+kx)$
$\ln L_2=\lim_{x\to 0} \frac{1}{x} (kx-\frac{k^2x^2}{2}+\frac{k^3 x^3}{3}-...)$
$\ln L_2=\lim_{x\to 0} (k-\frac{k^2x}{2}+\frac{k^3 x^2}{3}-...)$
$\ln L_2=k$
$L_2=e^k$

Since both converges to the same limit, therefore both limits are equivalent.

 

[Mark44]

Will this work as a proof?

$L_1=\lim_{x\to\infty} (1+\frac{k}{x})^x$
$\ln L_1=\ln[\lim_{x\to\infty} (1+\frac{1}{k})^x]$
$\ln L_1=\lim_{x\to\infty} \ln[(1+\frac{1}{k})^x]$$\ln(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-...$
$\ln(1+\frac{k}{x})=\frac{k}{x}-\frac{k^2}{2x^2}+\frac{k^3}{3x^3}-...$

$=\lim_{x\to\infty} x [\frac{k}{x}-\frac{k^2}{2x^2}+\frac{k^3}{3x^3}-...]$
$=\lim_{x\to\infty} [k-\frac{k}{2x^2}+\frac{k^2}{3x^3}-...]$
$=k$

Since
$\ln L_1=k$

Therefore
$L_1=e^k$

For the right hand side
$L_2=\lim_{x\to 0} (1+kx)^\frac{1}{x}$
$\ln L_2=\ln\lim_{x\to 0} (1+kx)^\frac{1}{x}$
$\ln L_2=\lim_{x\to 0} \ln (1+kx)^\frac{1}{x}$
$\ln L_2=\lim_{x\to 0} \frac{1}{x} \ln (1+kx)$
$\ln L_2=\lim_{x\to 0} \frac{1}{x} (kx-\frac{k^2x^2}{2}+\frac{k^3 x^3}{3}-...)$
$\ln L_2=\lim_{x\to 0} (k-\frac{k^2x}{2}+\frac{k^3 x^2}{3}-...)$
$\ln L_2=k$
$L_2=e^k$

Since both converges to the same limit, therefore both limits are equivalent.

That's an awful lot of work to prove something that's very straightforward. If $x \to \infty$ then $\frac 1 x \to 0$. Does that give you an idea of the substitution to do?

 

[blue_leaf77]

Will this work as a proof?

$L_1=\lim_{x\to\infty} (1+\frac{k}{x})^x$
$\ln L_1=\ln[\lim_{x\to\infty} (1+\frac{1}{k})^x]$
$\ln L_1=\lim_{x\to\infty} \ln[(1+\frac{1}{k})^x]$$\ln(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-...$
$\ln(1+\frac{k}{x})=\frac{k}{x}-\frac{k^2}{2x^2}+\frac{k^3}{3x^3}-...$

$=\lim_{x\to\infty} x [\frac{k}{x}-\frac{k^2}{2x^2}+\frac{k^3}{3x^3}-...]$
$=\lim_{x\to\infty} [k-\frac{k}{2x^2}+\frac{k^2}{3x^3}-...]$
$=k$

Since
$\ln L_1=k$

Therefore
$L_1=e^k$

For the right hand side
$L_2=\lim_{x\to 0} (1+kx)^\frac{1}{x}$
$\ln L_2=\ln\lim_{x\to 0} (1+kx)^\frac{1}{x}$
$\ln L_2=\lim_{x\to 0} \ln (1+kx)^\frac{1}{x}$
$\ln L_2=\lim_{x\to 0} \frac{1}{x} \ln (1+kx)$
$\ln L_2=\lim_{x\to 0} \frac{1}{x} (kx-\frac{k^2x^2}{2}+\frac{k^3 x^3}{3}-...)$
$\ln L_2=\lim_{x\to 0} (k-\frac{k^2x}{2}+\frac{k^3 x^2}{3}-...)$
$\ln L_2=k$
$L_2=e^k$

Since both converges to the same limit, therefore both limits are equivalent.

You can also do it that way, although as Mark hinted above, substitution may lead to significant reduction in the steps needed to do the proof.

 

[sooyong94]

That's an awful lot of work to prove something that's very straightforward. If $x \to \infty$ then $\frac 1 x \to 0$. Does that give you an idea of the substitution to do?

u=1/x?

 

[sooyong94]

I think I got the answer - do check if I commit any mistakes:

$x=\frac{1}{u}$, $u=\frac{1}{x}$
When $u \rightarrow \infty$, $x \rightarrow 0$

Then the limit can be rewritten as
$\lim_{u\to 0} (1+ku)^\frac{1}{u}$

Replacing u with x gives
$\lim_{x\to 0} (1+kx)^\frac{1}{x}$

Which is equal to the right hand side

 

[blue_leaf77]

Yes that's right.

 

[sooyong94]

Great! Can anyone mark this thread as solved? ;)