Splitting limit, where am I wrong? So we want to find this limit:
$$\lim \limits_{x \to 0}{\frac{e^{tgx}*\frac{1}{cos^2x}+e^{-tgx}*\frac{1}{cos^2x}-2}{6x^2}}$$
It can be done by geting $\frac{1}{cos^2x}$ from the denominator and making limit look like this:
$$\lim \limits_{x \to 0}{\frac{e^{tgx}+e^{-tgx}-2*cos^2x}{6x^2*cos^2x}}$$
Now we just pull $\frac{1}{cos^2x}$ outside, as it approaches 1 and get:
$$\lim \limits_{x \to 0}{\frac{e^{tgx}+e^{-tgx}-2*cos^2x}{6x^2}}$$
which we can Lopital 2 times and solve  using similar technique and we get the result $\frac{1}{2}$. $$$$
Now what what I am interested in is why couldn't we use the theorem that lets us split limits and say that:
$$\lim \limits_{x \to 0}{\frac{e^{tgx}*\frac{1}{cos^2x}+e^{-tgx}*\frac{1}{cos^2x}-2}{6x^2}}=\frac{\lim \limits_{x \to 0}{{(e^{tgx}*\frac{1}{cos^2x}+e^{-tgx}*\frac{1}{cos^2x}-2)}}}{\lim \limits_{x \to 0}{6x^2}}=\frac{\lim \limits_{x \to 0}{{(e^{tgx}*\frac{1}{cos^2x})}+\lim \limits_{x \to 0}{(e^{-tgx}*\frac{1}{cos^2x})}-{\lim \limits_{x \to 0}{2}}}}{\lim \limits_{x \to 0}{6x^2}}=
\frac{\lim \limits_{x \to 0}{{(e^{tgx})}*\lim \limits_{x \to 0}{(\frac{1}{cos^2x})}+\lim \limits_{x \to 0}{{(e^{-tgx})}*\lim \limits_{x \to 0}{(\frac{1}{cos^2x})}}-{\lim \limits_{x \to 0}{2}}}}{\lim \limits_{x \to 0}{6x^2}}=
\frac{\lim \limits_{x \to 0}{{(e^{tgx})}*1+\lim \limits_{x \to 0}{{(e^{-tgx})}*1}-{\lim \limits_{x \to 0}{2}}}}{\lim \limits_{x \to 0}{6x^2}}=
\frac{\lim \limits_{x \to 0}{{(e^{tgx}+e^{-tgx}-2)}}}{\lim \limits_{x \to 0}{6x^2}}=
\lim \limits_{x \to 0}{\frac{e^{tgx}+e^{-tgx}-2}{6x^2}}$$
And now we Lopital this 2 times and end up getting the result $\frac{1}{6}$.
So I want to know where is the mistake in this second procedure.
Thanks in advance!
 A: $$\lim_{x \to a} \Bigg(\frac{f(x)}{g(x)}\Bigg) = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}~~;~~\text{Provided: $\lim_{x \to a}g(x) \neq 0$}$$
As noted in the statement we only need to worry about the limit in the denominator being zero when we do the limit of a quotient.  If it were zero we would end up with a division by zero error and we need to avoid that.
In your case ; $\lim_{x \to a}g(x)$ i.e. $\lim_{x \to 0}6x^2 = 0$, Therefore you are ending up with wrong result.
A: The expression
$$\frac{\lim \limits_{x \to 0}{{(e^{\tan x}\frac{1}{\cos^2x}+e^{-\tan x}\frac{1}{\cos^2x}-2)}}}{\lim \limits_{x \to 0}{6x^2}}$$
is meaningless, as there is no number $\frac {0}{0}.$ Any implications you then "derive" from this will be of further damage to your mathematical health.
(Note the correct typesetting of the expression.)
A: Assuming what you're doing makes sense (it doesn't, I'm afraid), it is not clear why, when you are at
$$
\lim_{x\to0}e^{\tan x}\cdot\lim_{x\to0}\frac{1}{\cos^2x},
$$
you substitute the second limit with $1$ and leave the first unchanged. After all, also
$$
\lim_{x\to0}e^{\tan x}=1.
$$
So you'd get, at the numerator, $1+1-2=0$, which should make you suspect that you're dealing with a form $0/0$, where splitting the limit as the quotient of the limits is not admissible.
Note that the numerator is a continuous function, whose value at $0$ is $0$; similarly for the denominator.
This is a very common mistake. Consider
$$
\lim_{x\to0}\frac{1-\dfrac{\sin x}{x}}{x^2}.
$$
With your (wrong) method you could do
$$
\frac{\displaystyle\lim_{x\to0}1-\lim_{x\to0}\dfrac{\sin x}{x}}
{\displaystyle\lim_{x\to0}x^2}=
\frac{\displaystyle\lim_{x\to0}1-1}
{\displaystyle\lim_{x\to0}x^2}=\lim_{x\to0}\frac{1-1}{x^2}=0,
$$
but the correct value of the limit is $1/6$.
