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If $x_n$ is a sequence such that $x_n \neq 0$, and $x_n \rightarrow 0$, then prove:

$$\lim_{n \rightarrow \infty} \frac{\exp(x_n) - 1}{x_n} = 1.$$

where $\exp()$ is the exponential function. I was wondering if I would be able to use the fact that if $x_n \rightarrow 0$, then $\exp(x_n) \rightarrow 1$ in this proof, but I'm skeptical of using that because the numerator would approach $0.$ I just need a bit of guidance/maybe a hint. Thanks in advance!

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  • $\begingroup$ Can you use a taylor series expansion? If you could then expanding the exponential would make the answer very obvious. $\endgroup$
    – Kraigolas
    Mar 27 '20 at 20:14
  • $\begingroup$ @VividKraig We haven't "covered" Taylor series in my Real Analysis class yet, so I don't think I'd be able to use it here. $\endgroup$
    – user764255
    Mar 27 '20 at 20:18
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It is a direct consequence of the basic limit $$\lim_{t\to 0}\frac{e^t-1}{t}=1 $$ together with the sequential definition of a limit of a function which applied to the limit above gives you the precise statement that has to be shown.

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Observe that : $$ \left(\forall x\in\left[-1,1\right]\right),\ \left|\mathrm{e}^{x}-1\right|=\left|x\right|\int_{0}^{1}{\mathrm{e}^{xt}\,\mathrm{d}t}\leq \left|x\right|\int_{0}^{1}{\mathrm{e}^{t}\,\mathrm{d}t}=\left|x\right|\left(\mathrm{e}-1\right) $$

If $ \left(x_{n}\right)\in\mathbb{R}^{\mathbb{N}} $ such that $ x_{n}\underset{n\to +\infty}{\longrightarrow}0 $, then $ \left(\exists n_{0}\in\mathbb{N}\right)\left(\forall n\geq n_{0}\right),\ \left|x_{n}\right|\leq 1 $

Using the same trick : $$ \left(\forall n\geq n_{0}\right),\ \left|\frac{\mathrm{e}^{x_{n}}-1}{x_{n}}-1\right|=\left|\int_{0}^{1}{\left(\mathrm{e}^{x_{n}t}-1\right)\mathrm{d}t}\right|\leq\int_{0}^{1}{\left|\mathrm{e}^{x_{n}t}-1\right|\mathrm{d}t}\leq\left|x_{n}\right|\frac{\mathrm{e}-1}{2}\underset{n\to +\infty}{\longrightarrow}0 $$

Thus $$ \lim_{n\to +\infty}{\frac{\mathrm{e}^{x_{n}}-1}{x_{n}}}=1 $$

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  • $\begingroup$ thanks! I was wondering if this applies in general to a sequence $x_n$ such that $x_n \neq \alpha$, and $x_n \rightarrow \alpha$, so that: $\lim_{n \rightarrow \infty} \frac{\exp(x_n) - \exp(\alpha)}{x_n - \alpha} = \exp(\alpha).$ $\endgroup$
    – user764255
    Mar 29 '20 at 22:43
  • $\begingroup$ @John David factor out $ \exp{\alpha} $, denote $ y_{n}=x_{n}-\alpha $, then apply the previous result. $\endgroup$
    – CHAMSI
    Mar 30 '20 at 0:22

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