finding limit: result division by null I've got a question concerning this task:
$$\lim\limits_{x \rightarrow -{\frac{1}2}^+}{\frac{e^{2x-1}-1}{cos(\pi x)}}$$ 
So the limit value given in the solution is $-\infty$.

I am a bit confused about limit values and division by null in that context; So I know that in general $\lim\limits_{x \rightarrow -\frac{1}2}{\frac{1}{cos(\pi x)}}$ doesn't exist since the limit value is different for $\lim\limits_{x \rightarrow {-\frac{1}2}^-}$ and $\lim\limits_{x \rightarrow {-\frac{1}2}^+}$.
In my case ,though, it is possible because I'm only interested in $\lim\limits_{x \rightarrow -{\frac{1}2}^+}{cos(\pi x)}=0^+$ and therefore, with $\lim\limits_{x \rightarrow -{\frac{1}2}^+}{e^{2x-1}-1}={e^{-2}-1}^+$ that gives me $$\lim\limits_{x \rightarrow -{\frac{1}2}^+}{\frac{e^{2x-1}-1}{cos(\pi x)}}=\frac{{e^{-2}-1}^+}{0^+}$$ which is in that case $-\infty$, since the numerator is $&lt;0$, right?
So could I say that I can - in the context of limit values - divide by null if I'm only interested in the limit value from one side?

Thanks in advance.
It is not $e^{-2}$ but $e^0$. You have to use L'Hopital rule to get
$$
\lim\limits_{x \rightarrow -{\frac{1}2}^+}{\frac{e^{2x-1}-1}{\cos(\pi x)}}=
\lim\limits_{x \rightarrow -{\frac{1}2}^+}\frac{\frac{d}{dx}\left(e^{2x-1}-1\right)}{\frac{d}{dx} \cos(\pi x)}=
\lim\limits_{x \rightarrow -{\frac{1}2}^+}\frac{2 e^{2x-1}}{ -\pi \sin(\pi x)}=
-\frac{2}{\pi}
$$
It is not $e^{-2}$ but $e^0$. You have to use L'Hopital rule to get
$$
\lim\limits_{x \rightarrow -{\frac{1}2}^+}{\frac{e^{2x-1}-1}{\cos(\pi x)}}=
\lim\limits_{x \rightarrow -{\frac{1}2}^+}\frac{\frac{d}{dx}\left(e^{2x-1}-1\right)}{\frac{d}{dx} \cos(\pi x)}=
\lim\limits_{x \rightarrow -{\frac{1}2}^+}\frac{2 e^{2x-1}}{ -\pi \sin(\pi x)}=
-\frac{2}{\pi}
$$