Limit addition and multiplication question 
As x approaches one, the limit of f of x is 6, the limit of g of x is
  8, and the limit of h of x is 10. What is the limit as x approaches
  one of the function f + g times h?

So WLOG I just set $f(x) = 6/x$, $g(x) = 8/x$ and $h(x) = 10/x$. 
Then, $(f + g) * h$ = $\frac{10}{x} \cdot (\frac{6}{x} + \frac{8}{x}) = 140/x^{2}$ 
So $\lim_{x\to 1} 140/x^{2} = 140$. So my answer is $140$
But the correct answer is $86$. Why??
 A: 
Then, $(f + g) * h$ = $\frac{10}{x} \cdot (\frac{6}{x} + \frac{8}{x}) = 140/x^{2}$ 

But $(f+g) \cdot h \ne f+g \cdot h$ and they're asking:

the function f + g times h

Also: there's no need to invent functions satisfying the given limits; but if you want to, don't make it more difficult than necessary: the constant functions $6$, $8$ and $10$ do the trick!
A: We have that


*

*$\lim_{x\to 1} f(x)=6 \iff \forall \epsilon>0\quad \exists \delta_f>0\quad \forall x\quad 0<|x-1|<\delta_f \quad |f(x)-6|<\epsilon$

*$\lim_{x\to 1} g(x)=8\iff \forall \epsilon>0\quad \exists \delta_g>0\quad \forall x\quad 0<|x-1|<\delta_g \quad |g(x)-8|<\epsilon$

*$\lim_{x\to 1} h(x)=10\iff \forall \epsilon>0\quad \exists \delta_h>0\quad \forall x\quad 0<|x-1|<\delta_h \quad |h(x)-10|<\epsilon$
then $\forall \epsilon_1=3\epsilon>0$ indicating by $\delta=\min\{\delta_f,\delta_g,\delta_h\}$ we have that $\forall x\quad 0<|x-1|<\delta$
$$|f(x)+g(x)+h(x)-24|=|(f(x)-6)+(g(x)-8)+(h(x)-10)|\le $$
$$\le|f(x)-6|+|g(x)-8|+|h(x)-10|<3\epsilon=\epsilon_1$$
that is
$$\lim_{x\to 1} (f(x)+g(x)+h(x))=24$$
