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Find the limit as t approaches 0 from the right of $\frac{2}{t}-\frac{1}{e^t-1}$.

So, being that the function is undefined if we plug in zero, I've begun l'hopital's rule. $\frac{-2}{t^2}-(-e^{-t})$ where to from here. answer is infinity. this appears to be undefined already.

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    $\begingroup$ In the limit as $t\to0^+$, $\frac{2}{t}-\frac{1}{e^t-1}$ has the indeterminate form $\infty-\infty$. L’Hopital’s rule is only for indeterminate forms $\frac{0}{0}$ and $\frac{\infty}{\infty}$. When you have $\infty-\infty$, rewrite the difference as a quotient, look at the limit, and then use l’Hopital’s rule if appropriate. See OC-Sansoo’s answer. $\endgroup$ – Steve Kass Nov 12 '14 at 4:07
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$$ \lim_{t\to 0^+} \frac{2}{t}-\frac{1}{e^t-1}= \lim_{t\to 0^+} \frac{2e^t-2-t}{te^t-t} =\frac{0}{0}$$ So now we have $$ \lim_{t\to 0^+} \frac{\frac{d}{dx}[2e^t-2-t]}{\frac{d}{dx}[te^t-t]}= \lim_{t\to 0^+} \frac{2e^t-1}{te^t+e^t-1}= \frac{2-1}{0+1-1}= \frac{1}{0} =\infty$$

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HInts:

also you can use $e^x=1+x+\frac{x^2}{2!}+o(x^2)$

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  • $\begingroup$ what does the o signify? $\endgroup$ – aero26 Nov 12 '14 at 4:06
  • $\begingroup$ It means that high order of $x^2$ as $x\to 0$. $\endgroup$ – Paul Nov 12 '14 at 7:57
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We have by L'hospitale rule: $\dfrac{2}{t} - \dfrac{1}{e^t - 1} = \dfrac{2e^t-t-2}{te^t-t} \rightarrow \dfrac{2e^t-1}{e^t+te^t-1} \rightarrow \dfrac{1}{0} = +\infty$

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