Proving an identity for a free variable

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  • Thread starter Arman777
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  • #1
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Let us suppose we have a function such that

$$z = e^{1/ab} - 1$$

Where we have two free parameters, a and b.

Q1) Can we say that as ##b \rightarrow \infty##, ##z = 0##?

Or, since ##a## is a free parameter, there is always some value for ##a## such that ##z \neq 0## for ##b \rightarrow \infty## ?
 

Answers and Replies

  • #2
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If a does not depend on b then that limit is right. 1/(ab) will go to 0 for every fixed a if b goes to infinity, and the rest follows from the usual rules for limits.
 
  • #4
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If a does not depend on b then that limit is right. 1/(ab) will go to 0 for every fixed a if b goes to infinity, and the rest follows from the usual rules for limits.
I mean a is free so it does not depend on b. You can choose it any value you like. I first thougt the answer is 0. But then I asked to other people and they said we can set ##a = 1/b##, ##a = 1/3b## etc. But that seems strange because at that point theres no point for ##b \rightarrow \infty## ?
 
  • #6
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How can we proof that ?
##e^x=1 \Longleftrightarrow x=0## which can be seen from the definition ##e^x=\displaystyle{\sum_{k=0}^\infty \dfrac{x^k}{k!}}##.

##\dfrac{1}{ab}## is never zero.
But then I asked to other people and they said we can set ##a=1/b##, ##a=1/3b## etc.
... in which case ##a## is not independent from ##b##. A reasonable answer can only be given if ##a## and ##b## are specified. Here we have for constant ##a##
$$
\lim_{b \to \infty} e^{\frac{1}{ab}}=1
$$
 
  • #7
Office_Shredder
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A thing people would typically say is for any fixed a, the limit is zero as b goes to infinity.
 
  • #9
Stephen Tashi
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Let us suppose we have a function such that

$$z = e^{1/ab} - 1$$

Where we have two free parameters, a and b.

Q1) Can we say that as ##b \rightarrow \infty##, ##z = 0##?
What would you mean by saying that?

If ##f(x,y)## is a function of two variables, what definition would you use to explain the notataion "##lim_{x \rightarrow a} f(x,y)##"?

The only interpretation for that notation that I know about is that ##lim_{x \rightarrow a} f(x,y) = L(y)## where ##L## is a function of ##y##. The only interpretation for ##\lim_{x \rightarrow a} f(x,y) = k## is that the limit ##L(y)## is the constant function ##L(y) = k##.
 
  • #11
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For a=0 the expression is undefined anyway.
 

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