Can this limit be found without using L'Hôpital's rule or Taylor/Maclaurin series?-- $$L=\displaystyle\lim_{x \rightarrow 0} \dfrac{e^x-x-1}{x^2}$$

I came up to the right answer..just that the method is not foolproof.


Let $L$ be the limit. So,

$$L=\lim_{x \rightarrow 0}\dfrac{e^x-x-1}{x^2}$$ Now, let $x=2y$. So, $$L=\lim_{x \rightarrow 0} \dfrac{e^{2y}-2y-1}{4y^2}$$So, the limit can be rewritten as $$L=\lim_{y \rightarrow 0} \dfrac{e^{2y}-2e^y+1+2e^y-2y-2}{4y^2}$$ which is $$L=\dfrac{1}{4}\lim_{y \rightarrow 0} \left(\left(\dfrac{e^y-1}{y}\right)^2+2\dfrac{e^y-y-1}{y^2}\right)$$ Now comes what i was saying. What i did was i separated the limit across the two terms. We can do that only if the two limits exist individually and finitely. The first, i am sure, exists.But notice the second term is twice the limit we desire to find. So, this method is only applicable when the limit exists.If we do substitute the second term with 2L, we have $$L=\dfrac{1}{4} (1+2L)=L$$ Solving for $L$ we get $\boxed{L=\dfrac{1}{2}}$.

But as I said this method is not foolproof. Is there any?


We can use squeeze theorem .

For, $x$ sufficiently close to $0$ , following identity holds true:

$$ 1+x+\frac{x^2}{2}+x^3 \ge e^x \ge 1+x+\frac{x^2}{2}$$

These holds good, too but i have proved LHS part using taylor expansion, which is a crime to use in this problem. Any proof of LHS part, without expansion will do the work.

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