# Finding the limit without using L'Hôpital's rule or a Taylor/Maclaurin series.

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.

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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?

• – lab bhattacharjee Jul 12 '14 at 11:52
• @labbhattacharjee I do have a solution..don't know if it's right.. – user1001001 Jul 12 '14 at 12:01
• Why don't you add that to the Question & get it verified? – lab bhattacharjee Jul 12 '14 at 12:02
• Definitely show us your work @user157130 – Thomas Andrews Jul 12 '14 at 12:23
• – enzotib Jul 12 '14 at 12:23

For, $x$ sufficiently close to $0$ , following identity holds true:
$$1+x+\frac{x^2}{2}+x^3 \ge e^x \ge 1+x+\frac{x^2}{2}$$