Problem solving an exponential limit

[greg_rack]

Homework Statement

Which is the value of: $$\lim_{x\to 0^{-}}\frac{e^{kx}-1}{x}$$

Relevant Equations

none

I know it's probably an easy one, but I'm getting confused on how to treat that exponential numerator in order to escape from the indeterminate form $\frac{0}{0}$

 

[BvU]

Ever hear of Taylor series ?

Alternative: what is the definition of the derivative of $e^x$ at $x=0$ ?

 

[greg_rack]

Ever hear of Taylor series ?

Absolutely not and I'm pretty sure it's not covered by my high-school curriculum... not to excessively simplify, but isn't there another easier method? :)

 

[BvU]

I posted an alternative.

high-school curriculum

What does that say in relation to $e^x$ ?

 

[PeroK]

Absolutely not and I'm pretty sure it's not covered by my high-school curriculum... not to excessively simplify, but isn't there another easier method? :)

Why don't you suggest an idea from what you have learned in your curriculum?

 

[greg_rack]

I posted an alternative.
What does that say in relation to $e^x$ ?

I just had the idea of checking from notable(is that the right word?) limits, and
$$\lim_{x\to 0}\frac{a^x-1}{x}=ln(a)$$

 

[BvU]

Bravo!

(although I have no idea what you mean with 'notable limits' )

 

[greg_rack]

(although I have no idea what you mean with 'notable limits' )

I knew "notable" wasn't the right word. In Italian we call those "limiti notevoli"... try to check for a valid translation. These are limits which are apparently(at least for the knowledge of a high-schooler) impossible to solve and can be rewritten in more "comfortable" forms.

this is the table I'm relying on

I believe these are simply derivations done for those with the unnecessary mathematical knowledge to solve such limit-forms

 

[PeroK]

I knew "notable" wasn't the right word. In Italian we call those "limiti notevoli"... try to check for a valid translation. These are limits which are apparently(at least for the knowledge of a high-schooler) impossible to solve and can be rewritten in more "comfortable" forms.

View attachment 271174this is the table I'm relying on

I believe these are simply derivations done for those with the unnecessary mathematical knowledge to solve such limit-forms

Alternatively, if we let $f(x) = e^{kx}$, then what is $f'(0)$? Can you a) calculate this and b) express it as a limit?

PS another good way is using l'Hopital's rule.

 

[greg_rack]

Alternatively, if we let $f(x) = e^{kx}$, then what is $f'(0)$? Can you a) calculate this and b) express it as a limit?

PS another good way is using l'Hopital's rule.

$f'(0)$? Is it $f(0)$, considering $f(x) = e^{kx}$?

 

[BvU]

with the unnecessary mathematical knowledge to solve such limit-forms

You mean 'without the necessary ...', i.e. they come from out of the blue sky (dal cielo azurro ?)

 

[greg_rack]

You mean 'without the necessary ...', i.e. they come from out of the blue sky (dal cielo azurro ?)

Yes, I meant that of course... and yes, they come dal cielo azzurro for us :)

 

[kuruman]

I just had the idea of checking from notable(is that the right word?) limits, and
$$\lim_{x\to 0}\frac{a^x-1}{x}=ln(a)$$

That's a great idea. Now let $a=e^k$ and use this limite notevole.

 

[archaic]

since $x$ is going to $0^-$, you can approximate $e^{kx}$ by the linear function tangent to it at $x=0$; $g(x)=f'(0)(x-0)+f(0)=kx+1$. i think that this is what @PeroK was hinting at?

 

[PeroK]

since $x$ is going to $0^-$, you can approximate $e^{kx}$ by the linear function tangent to it at $x=0$; $g(x)=f'(0)(x-0)+f(0)=kx+1$. i think that this is what @PeroK was hinting at?

Explicitly, with $f(x) = e^{kx}$:
$$f'(0) = \lim_{h \rightarrow 0} \frac{f(h) - f(0)}{h} = \lim_{h \rightarrow 0} \frac{e^{kh} - 1}{h} = \lim_{x \rightarrow 0} \frac{e^{kx} - 1}{x}$$

 

[Office_Shredder]

It's not obvious to me that any of this is particularly relevant. Does the OP even know what a tangent line is?

 

[archaic]

It's not obvious to me that any of this is particularly relevant. Does the OP even know what a tangent line is?

hm.. i am not sure, but i think that the exponential and natural log functions are usually taught after the basic derivative chapters (which include the tangent to a function at a point)?

 

[kuruman]

It's not obvious to me that any of this is particularly relevant. Does the OP even know what a tangent line is?

Post #13 is certainly relevant because it is a followup on OP's post #6 and gives the correct result. Admittedly it's a cookbook approach but it is also within OP's grasp.

 

[PeroK]

It's not obvious to me that any of this is particularly relevant. Does the OP even know what a tangent line is?

The problem is that the only "pre-calculus" solution to this is to look it up in a table of known limits, which seems to be what the OP was expected to do.

A limit like that has got l'Hopital written all over it.

 

[kuruman]

The problem is that the only "pre-calculus" solution to this is to look it up in a table of known limits, which seems to be what the OP was expected to do.

A limit like that has got l'Hopital written all over it.

Or Taylor. The question is, has OP seen derivatives? We don't know, however we do know that OP has seen the limiti nottevoli.

 

[Mark44]

The problem is that the only "pre-calculus" solution to this is to look it up in a table of known limits, which seems to be what the OP was expected to do.

I agree.

A limit like that has got l'Hopital written all over it.

It sure does.

The question is, has OP seen derivatives?

Given that this thread is in the Precalc section, we shouldn't assume any knowledge of derivatives or limits or series.

 

[PeroK]

Given that this thread is in the Precalc section, we shouldn't assume any knowledge of derivatives or limits or series.

Not so easy when the question is about a limit!

 

[Mark44]

Not so easy when the question is about a limit!

Doh!

 

[greg_rack]

Not so easy when the question is about a limit!

By the way, that's the Italian questionable approach on precalc... and to those of you who were asking: no, I haven't yet seen calculus, but that's a matter of days since next week we'll start studying derivatives.
Evidently, the "limiti notevoli" are given in order to still solve(or just take confidence with?) more complex limits which require series or calculus in general without the need of knowing difficult concepts in advance in "early" classes.

PS: I'm sorry if I didn't take part on time to the discussion, but for some reason I wasn't getting notifications :)

 

[SammyS]

By the way, that's the Italian questionable approach on precalc... and to those of you who were asking: no, I haven't yet seen calculus, but that's a matter of days since next week we'll start studying derivatives.
Evidently, the "limiti notevoli" are given in order to still solve(or just take confidence with?) more complex limits which require series or calculus in general without the need of knowing difficult concepts in advance in "early" classes.

PS: I'm sorry if I didn't take part on time to the discussion, but for some reason I wasn't getting notifications :)

I don't know if you have solved this problem by now.

Look at Post #13 by @kuruman for help in using "limiti notevoli" to solve this.

 

[greg_rack]

I don't know if you have solved this problem by now.

Look at Post #13 by @kuruman for help in using "limiti notevoli" to solve this.

Yes, I do have solved it :)

 

[r731]

The numerator and denominator equal zero if you take the limit. Apply L'hopital's rule and then take the limit.

 

[epenguin]

Bravo!

(although I have no idea what you mean with 'notable limits' )

Standard, well-known.