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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?

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5 Answers 5

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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?

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    $\begingroup$ 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. $\endgroup$
    – Paramanand Singh ♦
    Commented Apr 19, 2017 at 2:58
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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 $$\ln(a) $$

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    $\begingroup$ 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 $\endgroup$
    – user12345
    Commented Apr 17, 2017 at 14:16
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    $\begingroup$ @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 $\endgroup$
    – user12345
    Commented Apr 17, 2017 at 14:48
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    $\begingroup$ @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. $\endgroup$
    – Paramanand Singh ♦
    Commented Apr 17, 2017 at 16:18
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    $\begingroup$ @ParamanandSingh you have to understand the question first! your answers may be correct but that is not what the question demands! $\endgroup$
    – user12345
    Commented Apr 19, 2017 at 9:18
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    $\begingroup$ @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. $\endgroup$
    – Paramanand Singh ♦
    Commented Apr 20, 2017 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?

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  • $\begingroup$ how did you get the third step? $\endgroup$
    – user12345
    Commented Apr 17, 2017 at 14:12
  • $\begingroup$ i have setted $$t=a^x-1$$ $\endgroup$
    – Dr. Sonnhard Graubner
    Commented Apr 17, 2017 at 14:13
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    $\begingroup$ thanks sonnhard. you meant the expression $\frac{a^x -1}{x} =\frac{t}{\frac{1}{\ln(a)}\ln(t+1)}$ got it. how do we proceed from there $\endgroup$
    – user12345
    Commented Apr 17, 2017 at 14:23
  • $\begingroup$ yes that is what i meant $\endgroup$
    – Dr. Sonnhard Graubner
    Commented Apr 17, 2017 at 14:24
  • $\begingroup$ As $x \to 0$, $t \to 0$; the new quotient in terms of t still results in the indeterminate form 0/0. What did you have in mind, @Dr. Sonnhard Graubner? $\endgroup$
    – electronpusher
    Commented Apr 19, 2017 at 1:13
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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 $$\frac{1+\ln(a)x - 1 + O(x^2)}{x} = \ln(a) + O(x) \to \ln(a)\text{ as } x \to 0$$

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

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