Expressing Derivative of Cable Tension Component I am looking at Hibbeler's section on a `Cable Subjected to a Distributed Load' in Engineering Mechanics - Statics, Ed13, as shown in the attached images.


I am getting stuck deriving equation (7-7). I derive the horizontal equilibrium equation just fine;
$$
|\vec{T} + \vec{\Delta T}| \cdot \cos(\theta + \Delta \theta) - |\vec{T}|\cdot \cos(\theta) = 0
$$
I rearrange to describe horizontal tension component as a function of $\Delta x$ (given $\Delta T$ and $\Delta \theta$ are ultimately functions of $\Delta x$);
$$
|\vec{T}|\cdot \cos(\theta) = |\vec{T} + \vec{\Delta T}|\cdot \cos(\theta + \Delta \theta)
$$
Then I express the differential;
$$
\frac{\Delta(|\vec{T}|\cdot \cos(\theta))}{\Delta x}
=
\frac{|\vec{T} + \vec{\Delta T}|\cdot \cos(\theta + \Delta \theta)}{\Delta x}
$$
Then when I go to express the derivative, I come up with a limit which does not exist. 
$$
\frac{d(|\vec{T}|\cdot \cos(\theta))}{dx}
=
\lim_{\Delta x \to 0}\frac{|\vec{T} + \vec{\Delta T}|\cdot \cos(\theta + \Delta \theta)}{\Delta x}
$$
Any pointers as to where my error is? Appreciate any advice offered. Thanks.
 A: First of all is better to use $T$ as a scalar and write: $$(
{T} + {\Delta T})\cdot \cos(\theta + \Delta \theta) - {T}\cdot \cos(\theta) = 0
$$ next expand the cosine: $$(
{T} + {\Delta T})\cdot (\cos(\theta)\cos( \Delta \theta)- \sin(\theta)\sin( \Delta \theta) )- {T}\cdot \cos(\theta) = 0
$$ and regroup: $$
{T}  \cos(\theta)(\cos( \Delta \theta) -1) + {\Delta T}\cdot \cos(\theta)\cos( \Delta \theta)- (T+ \Delta T) (\sin(\theta)\sin( \Delta \theta) ) = 0$$ now we can divide and take the limit: $$  \lim_{\Delta x  \rightarrow 0} 
{T}  \cos(\theta){(\cos( \Delta \theta) -1) \over \Delta \theta}   {\Delta \theta \over \Delta x} +  \lim_{\Delta x  \rightarrow 0} {\Delta T \over \Delta x}\cdot \cos(\theta)\cos( \Delta \theta)-  \lim_{\Delta x  \rightarrow 0} (T+ \Delta T) \sin(\theta){\sin( \Delta \theta)\over \Delta \theta}  {\Delta \theta \over \Delta x}= 0$$ I also introduced a couple of $ {\Delta \theta \over \Delta \theta}$ where needed.
Now, remember that also $\Delta \theta$ and $\Delta T$ go to $0$, so the first limit is $0$, in the second $\cos \Delta \theta$ goes to $1$ and in the third we can neglect $\Delta T$, we remain with: $${d T \over d x}\cos \theta-T \sin \theta {d \theta \over d x}=0$$ that is the expression we were looking for.
