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I am not sure why but I'm having a really tough time with the following problem:

Given: $$f(x) = \frac{11^x - 1}{x}$$ what should the extended function's value(s) be so that the function is continuous from the right and left?

Am I supposed to find the left-hand and right-hand limits and equate them to calculate the answer? I'm still in Calculus 1 so the L'Hopital rule doesn't apply yet.

EDIT: I haven't learned derivatives yet (or L'Hopital's rule) so is there any other way for me to solve this? Or did my professor throw in this problem prematurely?

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  • $\begingroup$ I think many books define that $e$ is the number such that $\lim_{x \to 0} \frac{e^x-1}{x}=1$. Use $11^x=e^{(\ln 11) x}$ to calculate the limit as $x \to 0$. $\endgroup$ – Ahmed S. Attaalla Sep 15 '18 at 5:53
  • $\begingroup$ Hints: what are the possible discontinuities? Which type are they? Could you resolve the discontinuity by some tools you have learnt? $\endgroup$ – xbh Sep 15 '18 at 5:54
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You are interested in evaluating

$$\lim_{x \to 0} \frac{11^x-1}{x}=\lim_{x \to 0} \frac{11^x-11^0}{x-0}=\frac{d(11^x )}{dx}\mid_{x=0}$$

You just have to evaluate the last term to know how to define $f$ at $x=0$.

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  • $\begingroup$ I haven't learned derivatives yet (or L'Hopital's rule) so is there any other way for me to solve this? Or did my professor throw in this problem prematurely? $\endgroup$ – blizz Sep 15 '18 at 16:24
  • $\begingroup$ what about series representation of exponential? $\endgroup$ – Siong Thye Goh Sep 15 '18 at 16:36
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The function $f(x)=\frac{11^x-1}{x}$ is defined everywhere except at $x=0$. In order for $f$ to be continuous at $x=0$, we need that:

$$f(0)=\lim_{x \to 0} f(x)$$

Note that:

$$f(x)= \ln 11 \left(\frac{e^{(\ln 11)x}-1}{(\ln 11) x } \right)$$

And since $\lim_{\text{whatever} \to 0} \frac{e^{\text{whatever}}-1}{\text{whatever}}=1$

.....

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  • $\begingroup$ The answer is ln(11) but I have no idea how it gets to that point... $\endgroup$ – blizz Sep 15 '18 at 13:14
  • $\begingroup$ Look how I’ve rewritten $f(x)$. The idea is that as $x \to 0$, we also have $(\ln 11) x \to 0$. So, $\lim_{x \to 0} \frac{e^{(\ln 11) x}-1}{(\ln 11) x}=\lim_{(\ln 11) x \to 0} \frac{e^{(\ln 11) x}-1}{(\ln 11) x}=1$ That is of course since $\lim_{\text{whatever} \to 0} \frac{e^{\text{whatever}}-1}{\text{whatever}}=1$.@blizz $\endgroup$ – Ahmed S. Attaalla Sep 15 '18 at 17:12

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