hobz Posted July 9, 2010 Share Posted July 9, 2010 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? Link to comment Share on other sites More sharing options...
DJBruce Posted July 9, 2010 Share Posted July 9, 2010 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] Link to comment Share on other sites More sharing options...
hobz Posted July 9, 2010 Author Share Posted July 9, 2010 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? Link to comment Share on other sites More sharing options...
DJBruce Posted July 9, 2010 Share Posted July 9, 2010 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. Link to comment Share on other sites More sharing options...
Dave Posted July 13, 2010 Share Posted July 13, 2010 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. Link to comment Share on other sites More sharing options...
hobz Posted July 14, 2010 Author Share Posted July 14, 2010 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. Link to comment Share on other sites More sharing options...
Dave Posted July 14, 2010 Share Posted July 14, 2010 Well hopefully h/h = 1, so I imagine its well defined! Link to comment Share on other sites More sharing options...
hobz Posted July 14, 2010 Author Share Posted July 14, 2010 Ah yes, and then L'Hopitâl doesn't apply since the limit does not converge to 0/0. Link to comment Share on other sites More sharing options...
the tree Posted July 14, 2010 Share Posted July 14, 2010 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]. Link to comment Share on other sites More sharing options...
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