# Finding limit without using limit

If we have to find the value of $$\lim_{x \to 0} \frac{e^x-1}{x}$$

I tried to solve this by using series i.e by expanding $e^x$ and got the result.

But if there is another method to solve this

• It is possible to use some upper and lower bounds for $e^x$ close to $x=0$, and use the squeeze theorem. – Scounged May 7 '16 at 12:31
• Can you apply l´hospital ? – callculus May 7 '16 at 12:32
• Can we use heine principle – Koolman May 7 '16 at 12:34
• What is heine principle ? – callculus May 7 '16 at 12:42
• @callculus heine principle states that if $lim f(x) = l$ where $x$ approaches to $a$ means whenever we take sequence of number $x1 ,x2 .....xn$ all lying in the domain and converging at a point $a$ then the corresponding value of $x$ $y=f(x1), f(x2) ,........f(xn)$ must converge to a finite number l . But the converse is not true. – Koolman May 7 '16 at 12:48

If one knows that $$\left( e^x\right)'=e^x,\quad x \in \mathbb{R},$$ then, using the definition, for $f$ derivable near $a$, $$\lim_{x \to a}\frac{f(x)-f(a)}{x-a}=f'(a)$$ yields

$$\lim_{x \to a}\frac{e^x-1}{x}=\lim_{x \to a}\frac{e^x-e^0}{x-0}=e^0=1$$ as wanted.

• One should remark that most derivations of $(e^x)^{\prime}=e^x$ use this limit. – Rene Schipperus May 7 '16 at 12:33
• But not necessarily. – Olivier Oloa May 7 '16 at 12:33
• Which proof did you have in mind ? – Rene Schipperus May 7 '16 at 12:35
• You may define $e^x$ as a power series or as the inverse function of $\ln$... – Olivier Oloa May 7 '16 at 12:36
• Yes that also a common way of doing things. Is $ln x$ defined the integral ? in which case its derivative is trivial to find. – Rene Schipperus May 7 '16 at 12:40

There is a simpler way to calculate the limit provided we know the following two facts:

• $e^{x} \geq 1 + x$ for all $x$.
• $e^{x + y} = e^{x}\cdot e^{y}$ for all $x, y$.

For the derivation here we need the first inequality $e^{x} \geq 1 + x$ to hold for all $x \in (-1, \infty)$ and not on whole of $\mathbb{R}$. Also we need the second property only for the specific case when $y = -x$ so that $e^{x}e^{-x} = 1$.

Both these facts can be established with some effort (see second half of the answer) via definition $$e^{x} = \lim_{n \to \infty}\left(1 + \frac{x}{n}\right)^{n}\tag{1}$$ I now prove that $(e^{x} - 1)/x \to 1$ as $x \to 0$. First we handle the case when $x \to 0^{+}$. The inequality $e^{x} \geq 1 + x$ shows that $$\frac{e^{x} - 1}{x}\geq 1\tag{2}$$ Further let $0 < x < 1$ and then we have the inequality $$e^{-x} \geq 1 - x$$ We can take reciprocals to get $$e^{x} \leq \frac{1}{1 - x}$$ Here we make use of the fact that $e^{x + y} = e^{x}e^{y}$. And this inequality gives us $$\frac{e^{x} - 1}{x} \leq \frac{1}{1 - x}\tag{3}$$ From equations $(2)$ and $(3)$ we get $$1 \leq \frac{e^{x} - 1}{x} \leq \frac{1}{1 - x}$$ for $0 < x < 1$ and letting $x \to 0^{+}$ and using Squeeze theorem we get $$\lim_{x \to 0^{+}}\frac{e^{x} - 1}{x} = 1\tag{4}$$ Note that the above limit implies that $e^{x} \to 1$ as $x \to 0^{+}$.

Next we handle the case when $x \to 0^{-}$. We put $x = -y$ to get $$\lim_{x \to 0^{-}}\frac{e^{x} - 1}{x} = \lim_{y \to 0^{+}}\frac{e^{-y} - 1}{-y} = \lim_{y \to 0^{+}}\frac{e^{y} - 1}{y}\cdot\frac{1}{e^{y}} = 1\cdot 1 = 1\tag{5}$$ and using equations $(4)$ and $(5)$ we have $$\lim_{x \to 0}\frac{e^{x} - 1}{x} = 1$$

For those who are interested in the proof of the properties of $e^{x}$ mentioned at the start of my answer, I provide an outline (with some details to be filled by the reader). Before we can proceed to establish the properties of $e^{x}$ it is necessary to show that the definition of $e^{x}$ via limit in $(1)$ makes sense.

It is easy to prove (and this is available in many textbooks and also on MSE) that the following limit $$\lim_{n \to \infty}\left(1 + \frac{1}{n}\right)^{n}\tag{6}$$ exists and is normally denoted by $e$. Moreover $2 < e < 3$. Using the existence of limit $(6)$ and a little amount of algebraic manipulation it is easy to prove that if $x$ is a positive integer then $$\lim_{n \to \infty}\left(1 + \frac{x}{n}\right)^{n} = e^{x}$$ For a real number $x$ we need some more work. For $x > 0$ it is easy to see via binomial theorem that sequence $(1 + (x/n))^{n}$ is increasing and further if $r$ is a positive integer with $r > x$ then $(1 + (x/n))^{n} < (1 + (r/n))^{n}$. Now $(1 + (r/n))^{n}$ is also increasing and tends to $e^{r}$ hence it follows that $(1 + (r/n))^{n} \leq e^{r}$ for all positive integers $n$. Hence the sequence $(1 + (x/n))^{n}$ is also bounded above by $e^{r}$ and thus tends to a limit. It follows that the limit $$F(x) = \lim_{n \to \infty}\left(1 + \frac{x}{n}\right)^{n}$$ exists for all $x \geq 0$ and moreover since the sequence is increasing the limit is not less than the value of sequence at $n = 1$. Thus we see that $$F(x) = \lim_{n \to \infty}\left(1 + \frac{x}{n}\right)^{n} \geq 1 + x\tag{7}$$ for all $x \geq 0$. Thus $F(x) > 0$ for all $x \geq 0$.

For $x < 0$ let us put $y = -x$ so that $y > 0$. The limit $$F(y) = \lim_{n \to \infty}\left(1 + \frac{y}{n}\right)^{n}\tag{8a}$$ exists and is positive. Further we know via Bernoulli's Inequality that if $n > |x|$ then $$1 - \frac{x^{2}}{n} \leq \left(1 - \frac{x^{2}}{n^{2}}\right)^{n} \leq 1$$ and hence on taking limit when $n \to \infty$ we get via Squeeze Theorem $$\lim_{n \to \infty}\left(1 - \frac{x^{2}}{n^{2}}\right)^{n} = 1\tag{8b}$$ Dividing $(8b)$ by $(8a)$ (and noting that $y = -x$) we see that the limit $$\lim_{n \to \infty}\left(1 + \frac{x}{n}\right)^{n}$$ exists and is positive. Thus we have proved that the limit $$F(x) = \lim_{n \to \infty}\left(1 + \frac{x}{n}\right)^{n}$$ exists for all $x$ and is positive and the definition $(1)$ is now justified. Moreover equation $(8b)$ shows that $F(x)F(-x) = 1$ for all $x$. Therefore we have $$F(-x) = 1/F(x) = \lim_{x \to \infty}\left(1 + \frac{x}{n}\right)^{-n}$$ and using binomial theorem for general index and assuming $0 < x < 1$ we can prove that $F(-x) \geq 1 - x$. Combined with equation $(7)$ it follows that $F(x) \geq 1 + x$ for all $x \in (-1, \infty)$ and this establishes the first property of $F(x)$. Also note that we have proved the specific case of second property namely $F(-x)F(x) = 1$ which was required in our solution, but it is easy to prove in general that $F(x + y) = F(x)F(y)$. For this we use the following lemma:

Lemma 1: If $x_{n}$ is a sequence of real or complex terms such that $n(x_{n} - 1)\to 0$ as $n \to \infty$ then $x_{n}^{n} \to 1$ as $n \to \infty$.

This lemma is proved in this answer. Now let us define the sequence $x_{n}$ by $$x_{n} = \dfrac{\left(1 + \dfrac{x + y}{n}\right)}{\left(1 + \dfrac{x}{n}\right)\left(1 + \dfrac{y}{n}\right)}$$ and it is easy to check that $n(x_{n} - 1) \to 0$ and therefore $x_{n}^{n} \to 1$ so that $F(x + y) = F(x)F(y)$ for all $x, y$.

• A well-founded approach. +1. – RRL May 8 '16 at 6:10

The sequence $\left(1+\frac{x}{n}\right)^n$ increases to $e^x$ thus setting $n=1$ we get

$$1+x\leq e^x$$ which gives

$$1\leq \frac{e^x-1}{x}$$

Similarly the sequence $\left(1+\frac{x}{n}\right)^{n+1}$ decreases to $e^x$

so for an arbitrary but fixed $n$,

$$e^x \leq \left(1+\frac{x}{n}\right)^{n+1}$$

And this gives $$\frac{e^x-1}{x} \leq \frac{1}{x}\left[\left(1+\frac{x}{n}\right)^{n+1}-1\right]$$ Now take the limit $x\to 0$ we get

$$1\leq \lim\limits_{x\to 0} \frac{e^x-1}{x} \leq \frac{n+1}{n}$$

Since $n$ was arbitrary the result follows.

• Can we use heine principle. – Koolman May 7 '16 at 12:42
• The squeeze (your heine) principal? – steven gregory May 7 '16 at 15:24
• @StevenGregory You are trying to squeeze my heine ? – Rene Schipperus May 7 '16 at 15:26
• There is an error when you take limits as $x \to 0$. You don't know beforehand that $(e^{x} - 1)/x$ tends to a limit so your last inequality does not make sense. It is best to use $e^{x} = \lim_{n \to \infty}(1 + (x/n))^{n}$ and then calculate limit of $(e^{x} - 1)/x$ as shown in this answer of mine: math.stackexchange.com/a/541330/72031 Your method can however be salvaged with more effort via careful $\epsilon, \delta$ arguments, but taking limits to obtain inequality directly does not work unless limits are guaranteed to exist. – Paramanand Singh May 8 '16 at 5:10
• @ParamanandSingh Oh you guys on stackexchange just LOOOVE to criticise, dont you ? You cant be polite about it, can you ? You cant say "what about this ?" or "how do you know ?" You just jump if with "There is an error". How arrogant. How rude. Anyway, from the two Bernoulli inequalities $1-x\leq e^{-x}$ and $1+y-x\leq e^{y-x}$ the monotonicity $\frac{e^x-1}{x} < \frac{e^y-1}{y}$ for $x<y$ follows by pure algebraic manipulation. SO THERE ! – Rene Schipperus May 8 '16 at 13:16

There are many properties of $e$ of the form "$e$ is the unique number such that...". Any of these can be taken as the def'n of $e$. We may define $e$ by $\int_1^e(1/x)\;dx=1.$ We can define $\ln (x)=\int_1^x=\int_1^x(1/y)\;dy\; ,$ for $x>0.$

Then $\ln (a b)-\ln a=\int_a^{a b}(1/y)\;dy.$ Putting $y=y' a$ in this integral, we obtain $\ln (a b)-\ln a=\int_1^b(1/y')\;dy'=\ln b$ for all positive $a,b$.

From this we can readily show that $e^x$ is the inverse of $\ln x,$ that is, $e^{\ln x}=x.$ We also have : The function $\ln x$ is 1-to-1 and continuous, and has no upper or lower bound .

So for $x\ne 0$ let $x=\ln (1+y)$ with $y\ne 0.$ Then $$(e^x-1)^{-1}\cdot x=((1+y)-1)^{-1}\cdot \ln (1+y)=y^{-1}\int_1^{1+y}(1/z)\;dz.$$ Since the function $f(z)=1/z$ is continuous in a neighborhood of $z=1,$ we have, by the fundamental theorem of calculus, $$\lim_{y\to 0}\; y^{-1}\int_1^{1+y}(1/z)\;dz=f(1)=1.$$