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Show that $\displaystyle\lim_{x\to0}\frac{e^x-1}{x}=1$

Letting $y=e^x-1\implies e^x=y+1\implies x=\log(y+1)$ the evaluation is easy.

But I can't understand how to express given function as a composition of two functions so that the following rule can be used limit operation can be done.

Let $A\subset\mathbb R,f:A\to\mathbb R,g:D\to\mathbb R$ such that $f(A)\subset D.$ Let $c$ be a limit point of $A$ and $\lim_{x\to c}f(x)=l.$ If $l\in D$ and $g$ is cont at $l$ then $\lim_{x\to c}(gf)(x)=g(l)$ and if $l\notin D$ but a limit point of $D$ then $\lim_{x\to c}(gf)(x)=\lim_{y\to l}g(y).$

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    $\begingroup$ There are many ways to solve this limit, all depending on what you want the starting point to be. How is $e^x$ defined for you? $\endgroup$ – vadim123 May 27 '13 at 14:19
  • $\begingroup$ As defined in the simple calculus as an inverse of $\log.$ I want to express given limit as $\lim_{x\to0}(gf)(x)$ and evaluate in terms of limiting values of $f,g.$ $\endgroup$ – Sriti Mallick May 27 '13 at 14:23
  • $\begingroup$ Please clarify what the last sentence in your question means. What is the "following rule"? $\endgroup$ – vadim123 May 27 '13 at 14:26
  • $\begingroup$ Let $A\subset\mathbb R,f:A\to\mathbb R,g:D\to\mathbb R$ such that $f(A)\subset D.$ Let $c$ be a limit point of $A$ and $\lim_{x\to c}f(x)=l.$ If $l\in D$ and $g$ is cont at $l$ then $\lim_{x\to c}(gf)(x)=g(l)$ and if $l\notin D$ but a limit point of $D$ then $\lim_{x\to c}(gf)(x)=\lim_{y\to l}g(y).$ $\endgroup$ – Sriti Mallick May 27 '13 at 14:29
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Start with $$e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!}=1+x+\frac {x^2}{2!}+\frac {x^3}{3!}+\ldots$$ This means that $$e^x - 1= x+\frac {x^2}{2!}+\frac {x^3}{3!}+\ldots$$ and that $$\frac{e^x - 1}x= 1+\frac {x}{2!}+\frac {x^2}{3!}+\ldots$$

I believe you can take it from here.

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Another way to do it is to define $f: \mathbb{R} \to \mathbb{R}$ by setting $f(x)=e^x$, then note that $f(0) = 1$, in that case the derivative of $f$ at the point $0$ can be given by the limit:

$$f'(0)=\lim_{x\to 0} \frac{f(x)-f(0)}{x-0}=\lim_{x\to 0}\frac{e^x - 1}{x}$$

In that case your limit is just $f'(0)$ which is $1$ since $f'(0) = e^0$. Of course this only works when you know that $f'(x) = e^x$. This helps you somehow?

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Use L'Hopital's rule:

$$\frac{d}{dx}(e^x -1 ) = e^x$$

And the denominator is $1$. Filling with the limit, gives $e^0/1 = 1$.

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Take $f:\mathbb R-\{0\}\to\mathbb R:x\mapsto e^x-1\\g:(-1,\infty)-\{0\}\to\mathbb R:x\mapsto\dfrac{x}{\log(x+1)}.$

Note $\forall~x\in\mathbb R,~e^x>0\text{ i.e. } e^x-1>-1\text{ and }e^x=1\iff x=0.\text{ So image}(f)\subset\text{domain}(g).$ Now $gf:\mathbb R-\{0\}\to\mathbb R:x\mapsto \dfrac{e^x-1}{x}.$

The given limit is $\displaystyle\lim_{x\to0}(gf)(x).$ $e^x-1$ being continuous at $0,\displaystyle\lim_{x\to0}f(x)=0,$ a limit point of the domain$(g)$ and $\displaystyle\lim_{x\to0}\frac{\log(1+x)}{x}=1(\neq 0)\implies\displaystyle\lim_{x\to0}\frac{x}{\log(1+x)}=1.$ Hence the given limit is $\displaystyle\lim_{x\to0}g(x)=1.$

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I would do this by using L'Hopitals rule. Because $\lim_{x \to 0} \frac{e^{x}-1}{x}= \frac{0}{0}$, we can take the the derivative of the numerator and denominator individually, then plug in 0 again to the new equation. After taking the derivative of the numerator and denominator we will have $\lim_{x \to 0}\frac{e^x}{1}$, and from here we see why the answer is $1$.

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  • $\begingroup$ Why did you remove your own question? :-) $\endgroup$ – mrs May 27 '13 at 18:54
  • $\begingroup$ I realized the question said "determine if this inequality is true, and if so prove that it is true via induction", and at that point I realized why the proof was not working inductively. The inequality was not true for all $n \epsilon \mathbb{N}$ $\endgroup$ – idk May 27 '13 at 20:13
  • $\begingroup$ So, here is your plus +1. I have kept it in my pocket for you. :-) $\endgroup$ – mrs May 28 '13 at 3:24

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