What is the limit of this thing

 

[math]

\lim_{n\rightarrow 0}\frac{a^n-1}{n}

[/math]

 

It has come up in an attempt of mine to find the derivative of [math]a^x[/math].

 

According to my derivatives table, the derivative of [math]a^x[/math] should be [math]a^x \ln(a)[/math], suggesting that the limit should converge to [math]\ln(a)[/math]. The question is, how do I arrive at this?

Start by evaluating the limit normally,

 

[math]

\lim_{n\rightarrow 0}\frac{a^n-1}{n}=\frac{a^0-1}{0}=\frac{1-1}{0}=\frac{0}{0}[/math]

 

The form [math] \frac{0}{0} [/math] is an indeterminate form, so we can apply L'Hopital. L'Hopital states that

 

if [math] \lim_{n\rightarrow c} \frac{f(n)}{g(n)}=\frac{0}{0} [/math]

 

then [math] \lim_{n\rightarrow c} \frac{f(n)}{g(n)}= \lim_{n\rightarrow c} \frac{f'(n)}{g'(n)} [/math]

 

So in your case we say:

 

[math]f(n)=a^n-1 \Rightarrow f'(n)= a^n(ln(a))[/math]

 

[math]g(n)=n \Rightarrow g'(n)=1[/math]

 

[math]

\lim_{n\rightarrow 0}\frac{a^n-1}{n}=\lim_{n\rightarrow 0}\frac{f'(n)}{g'(n)} = \lim_{n\rightarrow 0}\frac{(a^n)(ln(a))}{1}=a^0(ln(a))=(1)(ln(a))=ln(a) [/math]

In your derivation you use [math]f'(n)= a^n(\ln(a))[/math], but that is the very thing I am trying to prove. So how can the proof contain the original problem?

Sorry I misunderstood your question. In the case to prove that:

[math]\frac{d}{dx}a^x=ln(a)a^x[/math]

You can do it using implicit differentiation:

 

Let: [math] y=a^x[/math]

[math] y=a^x\Rightarrow ln(y)=ln(a^x)=x(ln(a))[/math]

 

Now lets take the derivative:

[math]\frac{y'}{y}= ln(a)(x') [/math]

 

Then multiply each side by y:

 

[math] y'=(y)ln(a)x' [/math]

 

Now we have that the derivative of [math]a^x[/math] is [math](y)ln(a)[/math], however, we can simplify this even more.

 

Remember in the beginning we let Let: [math] y=a^x[/math]. Therefore:

[math]\frac{dy}{dx}=(a^x)(ln(a)[/math]

 

As for arriving at this conclusion by evaluating the difference quotient. I do not know how to show the limit without using L'Hopital, which, would not make a valid proof.

Whilst you can use implicit differentiation, I think the OP is more interested in how you obtain this from the limit definition of the derivative. Having said this, I'm not 100% sure how to do it directly from the limit. The way you might approach it is to evaluate

 

[math]\lim_{x\to 0} \frac{e^x-1}{x} = e^x[/math].

 

This follows immediately from the Taylor series definition of [imath]e^x[/imath]. Then re-write [math]a^x = e^{x \ln a}[/math] and apply the chain rule.

Whilst you can use implicit differentiation, I think the OP is more interested in how you obtain this from the limit definition of the derivative. Having said this, I'm not 100% sure how to do it directly from the limit. The way you might approach it is to evaluate

 

[math]\lim_{x\to 0} \frac{e^x-1}{x} = e^x[/math].

 

This follows immediately from the Taylor series definition of [imath]e^x[/imath]. Then re-write [math]a^x = e^{x \ln a}[/math] and apply the chain rule.

 

Interesting. I will try and follow your suggestion, when I get some free time.

 

I've used

 

[math]

e = \lim_{x\rightarrow 0}\left(1+n\right)^{\frac{1}{n}}

[/math]

 

and substituted [math]a[/math] with that definition of [math]e[/math] in

 

[math]

\frac{\mathrm{d}}{\mathrm{d}x}\,a^x = a^x \cdot \lim_{h\rightarrow 0}{\frac{a^h-1}{h}}

[/math]

 

which eventually results in

 

[math]

\frac{\mathrm{d}}{\mathrm{d}x}\,e^x = e^x \cdot \lim_{h\rightarrow 0}{\frac{h}{h}}

[/math]

 

 

Edit:

Btw. Does the limit of [math]\lim_{h\rightarrow 0}{\frac{h}{h}}[/math] exist? Applying L'Hopitâl seems to yield 0/0 which is undefined.

Well hopefully h/h = 1, so I imagine its well defined!

Ah yes, and then L'Hopitâl doesn't apply since the limit does not converge to 0/0.

Whilst you can use implicit differentiation, I think the OP is more interested in how you obtain this from the limit definition of the derivative. Having said this, I'm not 100% sure how to do it directly from the limit.

I think, that the choice may be between using known results such as implicit differentiation - or by proving the limit from the [imath]\epsilon - \delta[/imath] definition of a limit.

 

To do that, rather than considering [math]\lim_{n \to 0} \frac{a^n - 1}{n}[/math] it'd probably be easier to use [math]\lim_{n \to \infty} n (a^{\frac{1}{n}} - 1)[/math].