How to compute $\lim\limits_{x \to 0} \frac{e^x-1}{x}$ without using derivatives? Every method I can think of gives me some indeterminate form.
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3Do you consider the power series expansion of $e^x$ to be "using derivatives"? – Ben Millwood Nov 18 '15 at 16:53
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$\displaystyle L=\lim_{x\rightarrow 0}\frac{e^{0+x}-e^0}{x}$. – JP McCarthy Nov 18 '15 at 17:01
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You are actually asking to compute $(e^x)'(0)$. – Nov 18 '15 at 17:03
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Are you allowed to use the fact that exponentials "go faster" than any polynomial? – Irregular User Nov 18 '15 at 17:03
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@IrregularUser Yes.Ben Millwood, I have never used power series expansion yet. – Bartosz S Nov 18 '15 at 17:08
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Then I assume that you've come across the sandwich/squeeze rule. Try and bound the limit from above and below. – Irregular User Nov 18 '15 at 17:13
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@IrregularUser Thanks, indeed I did use the rule. Still I have a hard time trying to find functions that would bound it to 1. – Bartosz S Nov 18 '15 at 17:24
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Set $\ e^x-1=y, $ so $\ x=\ln(1+y)$ and $\ y\to0$ for $\ x\to0$
Now you have: $$\ \lim_{x\to0}\frac{e^x-1}{x}=\lim_{y\to0}\frac{y}{\ln(1+y)}=\lim_{y\to0}\frac{1}{\frac{\ln(1+y)}{y}}=\lim_{y\to0}\frac{1}{\ln(1+y)^{\frac{1}{y}}}$$
Putting $\ \frac{1}{y}=t,$ for $\ y\to0, t\to\infty$ , so you have: $$\ \lim_{t\to\infty}\frac{1}{\ln(1+\frac{1}{t})^t}=\frac{1}{\ln(e)}=1$$
By definition of $\ e$.
Mosk
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Hint:
Take $x=\dfrac1k$ and use the definition of $e$:
$$\lim_{x\to0}\frac{e^x-1}x=\lim_{k\to\infty}k(e^{1/k}-1)=\lim_{k\to\infty}k\left(\lim_{nk\to\infty}\left(1+\frac1{nk}\right)^{nk/k}-1\right).$$
By the Binomial theorem, the inner limit expands as
$$1+\frac n{nk}+\frac{n(n-1)}{2n^2k^2}+\frac{n(n-1)(n-2)}{3!n^3k^3}\cdots,$$ which is bounded above by $$1+\frac1k+\frac1{2k^2}+\frac1{3!k^3}\cdots$$
Then $$1<k(e^{1/k}-1)<1+\frac1{2k}+\frac1{3!k^2}\cdots$$
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Hint: use the Taylor expansion for $e^x$ and cancel out the x in the denominator and then put x =0, to get 1 as the answer.
Nikunj
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One might argue that Taylor expansion is equivalent to knowing the derivatives... – Fabian Nov 18 '15 at 17:31
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