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Greetings to any and all that might read this. I am a 12-th grade student from Portugal and, here, it is not allowed for us to solve limits through L'Hôpital's Rule at this level (in fact, it isn't even taught, though we learn derivatives as well...), which places me (us) in a somewhat difficult position regarding the evaluation of some limits. Specifically, I've come across two particular limits that, somehow, have proven quite elusive to solve without recurring to L'Hôpital's Rule. Namely,

$$\lim _{x\to -2}\left(\frac{e^{x+2}-1}{\ln\left(7x+15\right)}\right)$$

and

$$\lim _{x\to 0}\left(\frac{\ln\left(x+1\right)-x}{x^2\left(1-x\right)} \right)$$

Through L'Hôpital's Rule, I've been able to determine that the first limit should yield $\frac{1}{7}$ and the second $-\frac{1}{2}$, but I haven't been able to reproduce this results without resorting to L'Hôpital's Rule, which is the way they demand the exercise to be solved. I've tried several variable substitutions, but all no avail; I might be missing something quite obvious, which is a mistake I often make, but I seem not to be able to solve them. In that sense, I was expecting that the Math Stack Exchange community could help me with this problem.

I thank you in advance for your answers and I apologize for any inconvenience my question might represent.

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  • $\begingroup$ Almost automatically, I would use Taylor series. in particular, that settles the second problem very quickly. Are Taylor series allowed? $\endgroup$ – André Nicolas Mar 2 '16 at 17:49
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    $\begingroup$ @André In response to your question, I believe Taylor series are not allowed, since they are not mentioned anywhere on our textbooks, neither has the teacher said anything about them... $\endgroup$ – NSFer 21 Mar 2 '16 at 19:02
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The first one is elementary if we set $f(x) = e^{x+2}, g(x) = \ln (7x+15).$ The expression then equals

$$\frac{(f(x)-f(-2))/(x-(-2))}{(g(x)-g(-2))/(x-(-2))}.$$

By definition of the derivative, the above $\to f'(-2)/g'(-2),$ which is easy to compute.

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    $\begingroup$ Thank you very much. I wouldn't have thought of making that function substitution (which, in essence, seems like a pretty good way to cirumvent the L'Hôpital's Rule restriction...). Sometimes, the way one is taught to do things (and the way one is more or less 'molded' to do them) gets in the way of thinking sufficiently "outside the box" (in regards to the usual wat one does thing) to actually be able to do them easily. So, thank you very much for your suggestion. $\endgroup$ – NSFer 21 Mar 2 '16 at 19:13
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$$ \lim _{x\to -2}\left(\frac{e^{x+2}-1}{\ln\left(7x+15\right)}\right)=\lim_{x\to 0}\left(\frac{e^x-1}{\ln (1+7x)}\right)\ . $$ Now multiply and divide by $7x$ $$ \lim_{x\to 0}\left(\frac{e^x-1}{7x}\frac{7x}{\ln (1+7x)}\right)\ . $$ Use the well-known limit $\lim_{\epsilon\to 0} \frac{\ln (1+\epsilon)}{\epsilon}=1=\lim_{\epsilon\to 0}\frac{e^\epsilon -1}{\epsilon}$, to conclude that indeed your limit is $1/7$. Such limits can be easily proven via Taylor expansions. If you haven't heard of Taylor expansions, and haven't seen these elementary limits before (they are on the same footing as $\lim_{x\to 0}\frac{\sin x}{x}=1$), then you'd need to tell us what your limited arsenal consists of and we'll try to see what we can do (but it's going to be tough/boring).

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  • $\begingroup$ Thank you very much. These limits (we call them something like "notable limits") are indeed taught. I just didn't have the clarity of mind to make that first step of converting $e^{x+2}$ to $e^{x}$ and $ln\left(7x+15\right)$ to $ln\left(7x+1\right)$... In what I did, I almost always ended up getting a $0\times\infty$ kind of indeterminate form... $\endgroup$ – NSFer 21 Mar 2 '16 at 19:08

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