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\begin{align} \lim_{x\to0} \frac{e^x-x-1}{x^2} &= \lim_{x\to0} \frac{e^x-1}{x^2} - \lim_{x\to0} \frac{1}{x} \\ &= \lim_{x\to0} \frac{e^x-1}{x}\lim_{x\to0} \frac{1}{x} - \lim_{x\to0} \frac{1}{x} \\ &= \lim_{x\to0} \frac{1}{x} - \lim_{x\to0} \frac{1}{x} \\ &= 0 \end{align}

Travis Willse
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oren revenge
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    The first equality already doesn't hold: Neither of the limits on the r.h.s. exist. – Travis Willse Mar 23 '16 at 15:10
  • See \http://math.stackexchange.com/questions/387333/are-all-limits-solvable-without-lh%C3%B4pital-rule-or-series-expansion – lab bhattacharjee Mar 23 '16 at 15:12
  • The error is, you can not use Limit Rules (substraction) when one of the limits does not exist.($\lim(A-B) \ne \lim(A) - \lim(B)$) – crbah Mar 23 '16 at 15:15
  • To expand on @Travis' comment: the method above is flawed, as you apply "normal" operations (sum, factorization, etc.) to quantities that do not exist: namely, $\lim_{x\to0} \frac{1}{x}$ -- in your proof, things go wrong as soon as you write this quantity). Basically, "anything can happen." – Clement C. Mar 23 '16 at 15:16
  • To illustrate one of the dangers of manipulating undefined quantities (here $\lim_{x\to 0}\frac{1}{x}$) as you did: from $1 = \frac{1+x}{x}-\frac{1}{x}$ (for any $x\neq 0$), you could write $$1 = \lim_{x\to 0} (1+x) \lim_{x\to 0} \frac{1}{x} - \lim_{x\to 0}\frac{1}{x} = \lim_{x\to 0}\frac{1}{x} - \lim_{x\to 0}\frac{1}{x} = 0$$which definitely isn't quite right. – Clement C. Mar 23 '16 at 15:26

3 Answers3

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Maybe you tried to apply this proposition:

If $\lim_{x\to a} f(x)=L_1$ and $\lim_{x\to a} g(x)=L_2$, then $\lim_{x\to a} (f(x)+g(x)) = L_1+L_2$.

But in this case, $\lim_{x\to 0}\frac{1}{x}$ does not converge.

choco_addicted
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The first claimed equality, $$\lim_{x\to0} \frac{e^x-x-1}{x^2} \color{red}{\stackrel{\textrm{(false)}}{=}} \lim_{x\to0} \frac{e^x-1}{x^2} - \lim_{x\to0} \frac{1}{x} ,$$ already doesn't hold: Neither of the limits on its r.h.s. exist.

To evaluate the limit:

Hint Apply l'Hopital's Rule (twice) or expand $e^x$ in a Maclaurin series.

Travis Willse
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The error starts here:

$$\lim_{x\to0} \frac{e^x-x-1}{x^2} $$ $$\color{red}{\neq\lim_{x\to0} \frac{e^x-1}{x^2} - \lim_{x\to0} \frac{1}{x} =\dots}$$

See Basic Limit Laws

Hint: Applaying L'Hopital twice gives $\frac 1 2 \lim\limits_{x\to 0} e^x$

3SAT
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  • i don't understand isnt $$ \lim_{x\to 0} \frac{e^x - 1}{x} = 1$$ ? – oren revenge Mar 23 '16 at 15:18
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    Uhm, that limit is certainly $1$, but that is not the issue in your computation, Orenrevenge. The problem is what you did when splitting the given into separate limits. Try L'Hospital on your very given limit and you should be fine – imranfat Mar 23 '16 at 15:21