Solving two specific limits without L'Hôpital's Rule 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.
 A: 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.
A: $$
\lim _{x\to -2}\left(\frac{e^{x+2}-1}{\ln\left(7x+15\right)}\right)=\lim_{x\to 0}\left(\frac{e^x-1}{\ln (1+7x)}\right)\ .
$$
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).
