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What steps are needed to evaluate the following?

$\lim_{x\to 1^-} \, e^{\frac{3}{1-x}}$

I know that the answer is $\infty$ but I don't know how to get there.

Thanks

P.S. I want to learn how to fish and not be given the fish so any generalized tips are welcome.

WXB13
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3 Answers3

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$$ Y=\lim_{x\to 1^-} \, e^{\frac{3}{1-x}} $$

$$ ln(Y)=\lim_{x\to 1^-} \, {\frac{3}{1-x}} ln(e)=\lim_{x\to 1^-} \, {\frac{3}{1-x}} $$

WeakLearner
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  • Aren't you then left with

    $$ ln(Y)=\lim_{x\to 1^-} , {\frac{3}{1-x}} $$

    instead of

    $$ Y=\lim_{x\to 1^-} , {\frac{3}{1-x}} $$

     – WXB13 Sep 13 '14 at 05:01
  • yeah that's what I wrote in my answer, not sure if its the best method but you are showing that since $ln(Y)=\infty$ then that should imply that what you were initially trying to solve is also $\infty$ – WeakLearner Sep 13 '14 at 06:08
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Show you that for every $\varepsilon > 0$ there is an $X < 1$ such that if $x \leq X$ then $$e^{\frac{3}{1-x}} > \varepsilon.$$ To see this, note that $$x \leq X \Rightarrow e^{\frac{3}{1-x}} > \varepsilon$$ is equivalent to $$e^{\frac{3}{1-x}} \geq e^{\frac{3}{1-X}} > \varepsilon,$$ so that the choice $$X := 1 - \frac{3}{\log 2\varepsilon}$$ suffices.

Yes
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  • Sorry but you've completely lost me. – WXB13 Sep 13 '14 at 05:04
  • @GaryWhite: Oh just wanted to convince you that why the given function is divergent as $x$ goes to 1 from the left :) – Yes Sep 13 '14 at 05:05
  • @GaryWhite: Another "intuitive" way to see the divergence may be: Let $\delta := 1-x$. Then, since $x < 1$, we have $ \delta > 0$ so that $$e^{\frac{3}{1-x}} = e^{\frac{3}{\delta}}$$ and hence, as $\delta \to 0$ we have $$e^{\frac{3}{\delta}} \to \infty.$$ – Yes Sep 13 '14 at 05:07
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Substituting $u=\frac{3}{1-x}$ yields that $$\lim_{x\to 1} \, e^{\frac{3}{1-x}} = \lim_{u\to\infty}e^u = \infty$$ since if you let $x\to1$ in the expression for $u$, then $u$ will approach $\infty$.

rehband
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