# What is the limit, If $\lim_{x\rightarrow \infty}(e^{-x} -1 )$ calculated from right side

what will be if we find the $\lim_{x\to \infty}(e^{-x} -1 )$ from right side such as

$$\lim_{x\to \infty}(e^{-x}-1) = \lim_{x\rightarrow \infty}\left( - x + \frac{x^2}2 - \frac{x^3}6 + \frac{x^4}{24}-\cdots\right)$$

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Hmmmm.... That seems to be an objectively terrible idea. Instead, examine the end behavior of $f(x)=e^{-x}-1$ as $x\to\infty$. Use the fact that $e^x$ is a strictly increasing positive function (for $x\in\Bbb R$) that tends to $\infty$ as $x$ tends to $\infty$. – Cameron Buie Sep 29 '12 at 4:13

1. It does not make sense to approach $\infty$ "from the right". What we mean by $\infty$ is a quantity which is bigger than every real number. If you draw the number line, approaching $\infty$ means we want to get as far to the right on that line as possible. The only way to approach $\infty$ is "from the left", so we instead just say $x \to \infty$, rather than $x \to \infty$ from the left.

2. LaTeX tip: if you use backslashes before your $lim$, you will get $\lim$. In other words, the output of $\lim$ will be $\lim$ rather than $lim$.

A few hints:

1. What happens to $e^x$ when $x \to \infty$? Therefore, what happens to $e^{-x} = \dfrac{1}{e^x}$?
2. What happens to $g(x) = 1$ when $x \to \infty$?
3. What can be said about $\displaystyle\lim_{x \to a} [f(x) +g(x)]$ if we know both $\displaystyle \lim_{x \to a} f(x)$ and $\displaystyle\lim_{x \to a} g(x)$?
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ok!! sir !!$\lim_{x\to \infty}(e^{-x}-1) =\lim_{x\to \infty}(e^{-x}) + \lim_{x\to \infty}(-1) = 0 -1 =-1$. – ram Sep 29 '12 at 4:44

$$\lim_{x\rightarrow \infty}\left( - x + \frac{x^2}2 - \frac{x^3}6 + \frac{x^4}{24}-\cdots\right)$$ is actually a double limit:

$$\lim_{x\rightarrow \infty}\left( - x + \frac{x^2}2 - \frac{x^3}6 + \frac{x^4}{24}-\cdots\right)=\lim_{x\rightarrow \infty} \lim_{m \to \infty} \sum_{k=1}^m \frac{(-x)^k}{k!}$$

Since in general double limits don't commute, you have to first calculate the inside limit in the RHS, and then you get the LHS..

Thus, if you calculate the RHS, you end up with the LHS.

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Sir!! Is we find other solution to it. I mean breaking the series to make other functions. – ram Sep 29 '12 at 4:32