I have worked out the following but need to prove it. lim x-->0 x^2 1-e^x^2 = -1 Can anyone help?
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It's of the indeterminate form $\displaystyle \displaystyle \frac{0}{0}$ so you can use L'Hospital's Rule.
Originally Posted by Arron I have worked out the following but need to prove it. lim x-->0 x^2 1-e^x^2 = -1 Can anyone help? For small $\displaystyle x$ $\displaystyle e^{x^2}=1-x^2+R(x)$ where $\displaystyle |R(x)|\le |x|^4/2$ So: $\displaystyle \frac{x^2}{1-e^{x^2}}=\frac{1}{-1+[R(x)/x^2]}$ etc. CB
If you let $\displaystyle t = x^2$, you could write your limit as $\displaystyle \displaystyle - \dfrac{1}{\lim_{t \to 0} \dfrac{e^t-1}{t}}$ with the limit now something you might recognize.
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