Proving an identity for a free variable

[Arman777]

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 $b \rightarrow \infty$ ?

 

[mfb]

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.

 

[fresh_42]

$z$ is never zero.

 

[Arman777]

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 there's no point for $b \rightarrow \infty$ ?

 

[Arman777]

$z$ is never zero.

How can we proof that ?

 

[fresh_42]

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 $a$
$$
\lim_{b \to \infty} e^{\frac{1}{ab}}=1
$$

 

[Office_Shredder]

A thing people would typically say is for any fixed a, the limit is zero as b goes to infinity.

 

[Arman777]

Yes $a$ should be constant

 

[Stephen Tashi]

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 $L(y) = k$.

 

[Svein]

A thing people would typically say is for any fixed a, the limit is zero as b goes to infinity.

_ as log as a≠0 ...

 

[mfb]

For a=0 the expression is undefined anyway.