Limit of the function $\lim_{x\to 0}(\frac{e^{-x}-1}{x})$

Question

I'm trying to solve this limit without the use of L'Hospital, but I'm doing something wrong. The limit should be:

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

My attempted proof: $$ \lim_{x\to 0}(\frac{e^{-x}-1}{x}) = \lim_{n\to \infty}(\frac{e^{-(\frac{1}{n})}-1}{\frac{1}{n}}) = \lim_{n\to \infty}(\frac{n \cdot ( e^{-(\frac{1}{n})}-1)}{1}) = \lim_{n\to \infty}(\frac{n \cdot ( e^{0}-1)}{1}) = \lim_{n\to \infty}(\frac{0}{1}) = 0 $$

I assume the mistake is that I've used the continuity of the $exp$ function.

Answer 1

A variation of Bernard's answer:

$$\frac{e^{-x}-1}x=\frac{1-e^x}{xe^x}=-\frac1{e^x}\cdot\frac{e^x-1}x\xrightarrow[x\to0]{}-\frac11\cdot(e^x)'_{x=0}=-e^0=-1$$

Answer 2

Hint:

This quotient is simply the rate of variation of the function $\mathrm e^{-x}$ starting from $x=0$, hence its limit is, by definition, the derivative of the function for $x=0$.

Answer 3

you are not applying L'Hospital correctly.

since the denominator contains only "X" after applying L'Hospital you should be getting only the derivation of the numerator, and from there it should be clear what the answer is.

this is what it should like after applying L'Hospital: See here

Answer 4

Without the use of derivatives, you can note that since $e = \lim_\limits{n \to \infty}\left(1+\frac{1}{n}\right)^n = \lim_\limits{n \to 0}(1+n)^{\frac{1}{n}}$, which is the usual definition of $e$, $e^{x} \sim 1+x$ and $e^{-x} \sim \frac{1}{1+x}$ for small values of $x$. Hence, the limit becomes

$$\frac{\frac{1}{1+x}-1}{x} = \frac{\frac{-x}{1+x}}{x} = -\frac{1}{1+x} \overset{x \to 0}{\to} -1$$

Alternatively, $e^{-x} \sim 1-x$ for small values of $x$ by the same definition, so you get $\frac{1-x-1}{x} = \frac{-x}{x} = -1$.

Addition: Your error is that you’re plugging in $n = \infty$ directly in your limit:

$$\lim_{n \to \infty} n\left(e^{-\left(\frac{1}{n}\right)}-1\right) \color{red}{\neq \lim_{n \to \infty} n\left(e^0-1\right)}$$

A simpler example demonstrating this mistake is the limit of $e$ itself:

$$e = \lim_{n \to \infty} \left(1+\frac{1}{n}\right)^n$$

If you treat $\frac{1}{n}$ separately from the rest of the limit, you may erroneously conclude that the limit is actually $(1+0)^n = 1$ because $\frac{1}{n} \to 0$ as $n \to \infty$.