3

So for evaluating limit of $\displaystyle\lim_{x\to0} \frac{e^{1/x} - 1}{e^{1/x} +1 } $ I used De l'Hospital rule, as conditions are being satisfied,

a) $f(x),g(x)\to \infty $

b) both are differential

And upon using L'Hospital rule I get $\displaystyle\lim_{x\to0} \frac{e^{1/x} (-1/x^2)}{e^{1/x} (-1/x^2)} $ , cancelling denominator with numerator i get 1.

But evaluating left hand and right hand limit separately shows that this function does not have limiting value at $x\to 0$

I don't understand what I am doing wrong ?

jjagmath
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Manu Sm
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    $e^{1/x}\not \to \infty$ when $x \to 0$ – jjagmath Apr 27 '23 at 11:05
  • They don’t go to infinity at 0 - the right limit is +infinity and the left is -infinity, so you need to be careful. Despite that, I do agree with 1 happening to be correct. If you divide the top and bottom by e^(1/x), you get the ratio of two expressions which have limit 1. – Eric Apr 27 '23 at 11:10
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    Maybe this helps https://math.stackexchange.com/q/2232656/505767 – user Apr 27 '23 at 15:11

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