How do I prove that both are equivalent limits

In summary, the conversation discusses how to prove that two limits, ##\lim_{x\to\infty} (1+\frac{k}{x})^x## and ##\lim_{x\to 0} (1+kx)^\frac{1}{x}##, are equivalent. One approach suggested is to use Taylor expansions, but another simpler approach is to substitute ##u=1/x##, which leads to a reduction in the number of steps needed for the proof. Both methods result in the two limits being equal.
  • #1
sooyong94
173
2

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'Hopitals 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?
 
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  • #2
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.
 
  • #3
sooyong94 said:

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'Hopitals 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.
 
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Likes QuantumQuest and blue_leaf77
  • #4
What substitution do you mean?
 
  • #5
sooyong94 said:
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?
 
  • #6
sooyong94 said:
What substitution do you mean?
blue_leaf77 said:
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}##
 
  • #7
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.
 
Last edited:
  • #8
sooyong94 said:
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?
 
  • #9
sooyong94 said:
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.
 
  • #10
Mark44 said:
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?
 
  • #11
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
 
  • #12
Yes that's right.
 
  • #13
Great! Can anyone mark this thread as solved? ;)
 

FAQ: How do I prove that both are equivalent limits

1. How do I know if two limits are equivalent?

Two limits are equivalent if they approach the same value as the independent variable approaches a certain value. This means that as the independent variable gets closer and closer to the specific value, both limits will get closer and closer to the same value.

2. What is the first step in proving equivalent limits?

The first step is to set up the equation for both limits and then simplify them as much as possible. This will help you see if the expressions are equivalent or not.

3. Can I use algebra to prove equivalent limits?

Yes, you can use algebraic manipulation to prove equivalent limits. This includes factoring, expanding, and combining like terms.

4. Are there any special cases when proving equivalent limits?

Yes, there are some special cases where proving equivalent limits can be more challenging. These include limits involving trigonometric functions, logarithmic functions, and piecewise functions.

5. How can I check my work when proving equivalent limits?

You can check your work by plugging in different values for the independent variable and seeing if both limits approach the same value. You can also use graphing software to visually see if the two functions have the same behavior near the specific value.

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