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The questions is: Evaluate $$\lim_{x\to 0} \frac{a^x -1}{x}$$ without applying L'Hopital's Rule.

Does this question fundamentally same as asking if the $\lim_{x\to 0} \frac{a^x -1}{x}$ exists? rather than straightway asking to find the limit. That means are questions (1) proving if the limit of a function exists and (2) asking what is the limit of that function, essentially same question?

user12345
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5 Answers5

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Hint: if you write $f(x) = a^x$, then the given limit has the form $$\lim_{x \to 0} \frac{f(x) - f(0)}{x-0}.$$ Does that look familiar?

Umberto P.
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    This implicitly assumes that $(a^{x}) '=a^{x} \log a$. The differentiation formula can be proved either by using the limit in question or by using chain rule. In both cases it depends on derivative of $e^{x} $ in which case this is same as the answer by Salahamam_Fatima. – Paramanand Singh Apr 19 '17 at 02:58
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we have

$$a^x=e^{x\ln(a)} $$ and

$$\lim_{t\to 0}\frac {e^t-1}{ t} =1$$

the limit is $$\ln(a) $$

hamam_Abdallah
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    to know $e^t-1\sim t ;;(t\to 0).$ is another requirement. That itself requires a proof. May be students need more simplified answer – user12345 Apr 17 '17 at 14:16
  • @user12345: one has to start from somewhere. This limit involving $e^{t} $ is one of those standard limits which are easily obtained via definition of $e^{t} $. It is similar to knowing and using $\lim_{x\to 0}\dfrac{\sin x} {x} =1$ used for all limits involving circular functions. Equivalently one can reduce the problem to limit involving logarithm function and then you need to know that $\lim_{x\to 0}\dfrac{\log(1+x)}{x}=1$. – Paramanand Singh Apr 17 '17 at 14:28
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    @ParamanandSingh $\lim_{x\to 0}\dfrac{\log(1+x)}{x}= \ln e=1$ is straightforward by definition of $e$ so nothing to calculate there, I dont mind if the expression simplifies to that. $\lim_{x\to 0}\dfrac{\sin x} {x} =1$ is a finding which is not by definition so one cannot just apply that limit in the question which does not even allow L'Hospital's Rule. Else using L'Hospital's Rule can solve it in one step – user12345 Apr 17 '17 at 14:48
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    @user12345: I think you need not adopt such a naive attitude. The theory of exponential and logarithmic functions can be developed from many fronts and each specific approach has its merits and challenges. Thus what may appear as an immediate consequence of definition may also appear as significant theorem if seen from a different perspective. Also if you are willing to take the log limit as fundamental then using $t=\log(1+x)$ we get $x=e^{t} - 1$ so that both the limits are immediate consequences of the other. – Paramanand Singh Apr 17 '17 at 16:18
  • @Paramanand Singh In this context, where L'Hospital's Rule is not allowed, a "naive attitude" seems to be a requirement of the question. I agree with user12345 that using the limit you suggest in your answer seems to be "cheating" rather than starting from "fundamental principles" (or those considered fundamental in this context, which is open to interpretation). – electronpusher Apr 19 '17 at 01:08
  • @electronpusher: who is using L'Hospital's Rule here? The solution is just using the definition of $e^{t} $ and it's consequences. Either you are not familiar with the theory of these functions or you have some deep misunderstanding of these things. I think it is better that you clarify what definitions you have in mind for symbols $e, e^{x}, \log x, a^{b} $. – Paramanand Singh Apr 19 '17 at 02:48
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    @ParamanandSingh you have to understand the question first! your answers may be correct but that is not what the question demands! – user12345 Apr 19 '17 at 09:18
  • @Paramanand Singh I concede that I do not have a degree in mathematics. However, I have studied these things quite a bit in my spare time. My understanding is that the best definition of $e^x$ is via its Taylor series expansion. Exponential functions of arbitrary base can be defined as transformations of $e^x$. The operation of exponentiation (as present in $a^b$) can also be defined in terms of $e^x$. The natural logarithm can be defined as the inverse function of the exponential function, or equivalently the solution $w$ to $e^w=z$. All other logs are proportional to the natural log. – electronpusher Apr 20 '17 at 14:42
  • These are the definitions I have in mind. Personally, I also think the limit definition of e is meaningful: $$e^x=\lim_{n \to \infty}(1+x/n)^n While I personally consider this a valid fundamental definition, I do not see reason for the limit you use in your answer to be considered a fundamental principle, to be used without proof. – electronpusher Apr 20 '17 at 14:46
  • @Paramanand Singh I have read your proof here (https://math.stackexchange.com/a/541330/398916) that the limit you use in your answer above follows from the limit definition of e. It might be a good idea to include the link to your proof in your answer here. In that case you'll have my vote. – electronpusher Apr 20 '17 at 15:02
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    @electronpusher: even I don't have a degree in math, and that part is actually immaterial here. If you define $e^{x} $ via Taylor series then you see that it's derivative at $0$ is coefficient of $x$ and hence it is $1$. Thus you get $\lim_{x\to 0}\dfrac{e^{x}-1}{x}=1$. So it is an immediate consequence of your definition without any use of L'Hospital's Rule. – Paramanand Singh Apr 20 '17 at 15:43
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set $$t=a^x-1$$ then we have $$x=\frac{1}{\ln(a)}\ln(t+1)$$ and you will get $$\frac{t}{\frac{1}{\ln(a)}\ln(t+1)}$$ can you finish?

Dr. Sonnhard Graubner
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Another way would be by Taylor expanding $$a^x = e^{\ln(a)x} = 1 + \ln(a)x + O(x^2)$$

And we get $$\frac{1+\ln(a)x - 1 + O(x^2)}{x} = \ln(a) + O(x) \to \ln(a)\text{ as } x \to 0$$

mathreadler
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Evaluating the limit $\mathop {\lim }\limits_{x \to 0} \frac{{{a^x} - 1}}{x}$ without applying the L'Hospital's Rule.

I think that one of best way for evaluating of this limit is using of the power series expansion for the function $a^x, (a > 0)$ about the point $x = 0$. Namely, we can write that $$ \begin{align*} a^x &= \sum_{k = 0}^\infty \left( x \ln(a) \right)^k \\ &= 1 + x \ln(a) + x^2 \ln(a)^2 + x^3 \ln(a)^3 + \cdots \\ &= 1 + x \ln(a) \left( 1 + x \ln(a) + x^2 \ln(a)^2 + x^3 \ln(a)^3 + \cdots \right) \\ &= 1 + x \ln(a) \sum_{k = 0}^\infty \left( x \ln(a) \right)^k \\ &= 1 + x \ln(a) \lim_{n \to \infty} \sum_{k = 0}^n \left( x \ln(a) \right)^k \\ &= 1 + x \ln(a) \lim_{n \to \infty } \frac{1 - \left(x \ln(a) \right)^{n + 1}}{1 - x \ln(a)} \end{align*} $$ by the sum of the geometric sequence.

So, we have got

$$a^x = 1 + x \ln(a) \lim_{n \to \infty } \frac{1 - \left( x \ln(a) \right)^{n + 1}}{1 - x \ln(a)}.$$

Substitution of this expression in $\lim_{x \to 0} \frac{a^x - 1}{x}$ gives us the following:

$$ \begin{align*} \lim_{x \to 0} \frac{a^x - 1}{x} &= \lim_{x \to 0} \frac{1 + x \ln(a) \lim_{n \to \infty} \frac{1 - \left( x \ln(a) \right)^{n + 1}}{1 - x \ln(a)} - 1}{x} \\ &= \lim_{x \to 0} \frac{x \ln(a) \lim_{n \to \infty } \frac{1 - \left( x \ln(a) \right)^{n + 1}}{1 - x \ln(a)}}{x} \\ &= \ln(a) \lim_{x \to 0} \left( \lim_{n \to \infty } \frac{1 - \left( { \ln(a)} \right)^{n + 1}}{1 - x \ln(a)} \right) \\ &= \ln(a) \lim_{n \to \infty } \left( \lim_{x \to 0} \frac{1 - \left( x \ln(a) \right)^{n + 1}}{1 - x \ln(a)} \right) \\ &= \ln(a) \frac{1 - 0}{1 - 0} \\ &= \ln(a). \end{align*} $$

So, $\mathop {\lim }\limits_{x \to 0} \frac{{{a^x} - 1}}{x}= \ln(a)$.

Sharif E. Guseynov
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