# Geometrical proof of the limit $\lim_{(x \to 0)}\left(\frac{e^x-1}{x}\right)=1$ using sandwich theorem.

I am studying about sandwich theorem and its applications by deriving some well-known limits such as this- $$\lim_{(x \to 0)}\left(\frac{e^x-1}{x}\right)=1$$ while I found some proofs of this result by first defining $e$ and then using that definition such as here(Proof of $f(x) = (e^x-1)/x = 1 \text{ as } x\to 0$ using epsilon-delta definition of a limit)(which I agree sounds a lot easier because if one is using the Taylor series expansion of $e^x$ then it becomes very easy)but my book tries to do this in a different manner by using this inequality $$\frac{1}{1+|x|}≤\left(\frac{e^x-1}{x}\right)≤ 1 + (e – 2) |x|$$(holds for all $x$ in $[–1, 1]-[0]$)

and then just using sandwich theorem the limit is easily calculated , but the book does not explain as to from where this inequality came from.And I am not able to get it by myself ,as I am not able to see how can this result be so obvious, and even though the graph does make it a bit clear(which I have attached below) still I am not able to get the given inequality(any hints there?)

I tried to search it on this site but couldn't find it,but still I found some very neat applications of sandwich theorem such as here ( How to prove that $\lim\limits_{x\to0}\frac{\sin x}x=1$?) so which makes me think that maybe this inequality can be derived easily by looking at it's geometrical interpretation such as in the link given.

So can someone please help me in understanding the geometrical meaning of this inequality (just like in the limit given in the link above) or help to derive it using its geometrical implications?

1)And even if that is not possible can someone help me in understanding the inequality intuitively because I honestly haven't got any idea as to how such a weird looking inequality can be related to the given limit,

2)And how that inequality is derived ?

3)Also what can possibly be the motivation behind this complicated inequality for deriving this limit, are there other such wierd inequalities also for finding this limit?

• I've been trying to help a friend study for calc one/two review, and we came up across this very problem in deriving the derivative to $e^x$. So I want to know the answer too! – Alan Mar 5 '16 at 19:19
• What definition of $e^x$ are you using? – zhw. Mar 6 '16 at 18:36
• @zhw none in particular my book has not first defined $e^x$ and all that instead has directly written the inequality ...you can use any definition you like...but as I pointed out in the question I have already seen some proofs of this kind(that is first defining $e^x$ and then using it again) instead I just want to know how the inequality in question is derived.... – Freelancer Mar 6 '16 at 19:04
• @zhw. so if you are also going to give such kind a proof then I would just like to point out that there are already many such proofs on this site...and I have already gone through them.. But they don't seem to help me in this particular question... – Freelancer Mar 6 '16 at 19:04
• @Freelancer: Unless you have a clear definition of $e^{x}$ it is impossible to derive any inequalities dealing with $e^{x}$. Unfortunately many calculus textbooks don't define these things and just assume that the student will somehow get hold of these things. The reason textbooks don't define $e^{x}$ is because it is hard! (especially for beginners in calculus). See paramanands.blogspot.com/2014/05/… and next few posts for definition of $e^{x}$ and proof of $(e^{x} - 1)/x \to 1$ as $x \to 0$. – Paramanand Singh Mar 7 '16 at 11:56

Just playing around to see what happens.

If $e(x+y) =e(x)e(y)$ with $e'(0) = 1$, then $e(x+h) =e(x)e(h)$ so $e(x+h)-e(x) =e(x)(e(h)-1)$ so, letting $h \to 0$ and using $e(0) = 1$, $e'(x) =e(x)e'(0) =e(x)$.

Therefore, $e(x)-1 =\int_0^x e(t) dt$.

Since $e(x) > 0$ for all $x$, for $x > 0$, from $e(x) =1+\int_0^x e(t) dt$, we get sequentially, $e(x) > 1$, $e(x) > 1+x$, $e(x) > 1+x+\frac{x^2}{2}$, so that, by induction, for any $n \ge 0$, $e(x) \gt \sum_{k=0}^n \frac{x^k}{k!}$.

For an upper bound on $e(x)$, assume that $0 < x < 1$ (since there is no polynomial bound for all $x$).

Let $f(x) = e(x)(1-x)$. $f(0) = 1$. $f'(x) =e'(x)(1-x)-e(x) =e(x)(1-x)-e(x) =-xe(x) < 0$ so $f(x)$ is decreasing for $x > 0$. Therefore $f(x) < f(0) = 1$, so $e(x) < \frac1{1-x}$.

To get a linear bound on $e(x)$, suppose that $0 < x < a < 1$. We want a $c > 1$ such that $\frac1{1-x} \le 1+cx$ for $0 < x \le a$.

This is $1 \le (1-x)(1+cx) =1+x(c-1)-cx^2$ or $cx \le c-1$ or $c(1-x) \ge 1$ or $c \ge \frac1{1-x}$.

Since $x \le a$, $\frac1{1-x} \le \frac1{1-a}$, so $c =\frac1{1-a}$ works.

Therefore $e(x) \le 1+\frac{x}{1-a}$ for $0 < x < a < 1$.

I'll leave it at this.

This is a response to OP's comments which is a bit long for a comment.

As I iterated in my comments to the question, the evaluation of limit of $(e^{x} - 1)/x$ as $x \to 0$ crucially depends on the definition of the symbol $e^{x}$. There are multiple approaches to define $e^{x}$ (which are provided here, here and here). One of these approaches is based on defining logarithm as an integral and treating exponential function as its inverse.

OP raises a further doubt about some sort of circularity involved here. His argument is that the concept of integrals is itself dependent on that of limits and hence something based on integrals should not be used as a basis to establish a certain limit. However it should be made clear that there is absolutely no circularity involved if we define $$\log x = \int_{1}^{x}\frac{dt}{t}$$ and further define $e^{x} = y$ if $x = \log y$. This is because the definition of $\log x$ as an integral is not dependent on any particular properties or features of $e^{x}$ (in particular the limit of $(e^{x} - 1)/x$ as $x \to 0$. The definition of $\log x$ is based on the properties of function $1/x$ and the concept of definite integrals as a limit of sum (thus the definition of $\log x$ is based on concept of limits).

On the other hand if one wishes to avoid the integrals (because from OP's point of view integrals come much later in study of calculus compared to limits) then we can directly use limits to define $e^{x}$ as follows $$e^{x} = \lim_{n \to \infty}\left(1 + \frac{x}{n}\right)^{n}$$ and using this definition we can easily prove that $\lim\limits_{x \to 0}\dfrac{e^{x} - 1}{x} = 1$.

Note however that such proofs can be daunting both for students (and teachers alike to explain) in a first course of calculus (meaning students are of age 16-17 years). It is preferable to be a bit honest and instead of hand waving introduce the functions $\log x, e^{x}, a^{x}$ with a list of their properties including the following standard limits $$\lim_{x \to 0}\frac{\log(1 + x)}{x} = 1 = \lim_{x \to 0}\frac{e^{x} - 1}{x},\, \lim_{x \to 0}\frac{a^{x} - 1}{x} = \log a$$ and mention that these properties can/will be proved in a later course on advanced calculus/real analysis.

However I have seen that most instructors almost always prefer hand waving instead of deferring the rigorous approach to later courses. This is one of reasons many students feel confused (OP seems to be a genuine victim of such intellectual fraud because his textbook author has the audacity to prove the limit of $(e^{x} - 1)/x$ via squeeze theorem without a definition of $e^{x}$).

• @Freelancer: I think it is better to keep the question intact. Your question is not wrong. It shows your confusion regarding these topics and perhaps it can help some other students like you. About accepting an answer it is up to you to choose any answer which you like most and accept it. – Paramanand Singh Mar 8 '16 at 4:29

i am not sure if we are allowed to use integrals, if we are, then using the mean value theorem for integrals, i can write $$\frac{1-e^x}{x} = \frac 1x\int_0^x e^t \, dt = e^{\theta x} \text{ for some 0 < \theta < 1}.$$ we now use the fact that $e^{\theta x}$ is between $e^x$ and $1$ to conclude the limit as $x$ tend to $0$ is $1$ using the squeeze theorem.

p.s.: i can prove the inequality $$f(x) = \frac x{x+1} \le e^x - 1 = g(x)$$ by showing that $g$ is concave up and $f$ is concave down. we also have one common tangent $y = x$ at $x = 0, y = 0$ which gives us $$\frac x{x+1} \le x \le e^x - 1$$ equality iff $x = 0.$

i can show the other inequality too. we will only need to show for $x > 0$ that $$h(x) = e^x-1 \le x + bx^2 = j(x)$$ we know that the graph of $h$ will cross the graph of $j.$ we just choose $b$ so that they cross at $x = 1.$ that means $e-1 = 1 + b$ which gives you the value for $b = e-2.$

• You are using the fact that $(e^t)'=e^t.$ But if we can use that, then we're already done, as the expression is the limit of difference quotients that converge to $(e^t)'(0) = e^0 = 1.$ – zhw. Mar 6 '16 at 18:18
• @Zhw, i wanted to use the squeeze theorem. if all we wnated is the limit, then of course you are right. – abel Mar 6 '16 at 18:20
• Even so, using $(e^t)'=e^t$ to calculate the derivative of $e^t$ at $t=0$ is wrong, no matter which method you try to use. – N. S. Mar 6 '16 at 18:29