Question 637331
I'm going to assume that your problem is
{{{lim(h->0, ((5x+5h+1)-(5x+1))/h)}}}
(Please use parentheses to make your expressions clearer.)<br>
Let's simplify the fraction. In the numerator, the 5x's and the 1's cancel out:
{{{lim(h->0, (5h)/h)}}}<br>
Next we could cancel the h's, <i>but only if h is not zero!</i> An important thing to understand about this limit is that we are interested in what happens when h gets infinitely close to zero <i>but not actually zero.</i> Since h is never actually zero, it is OK to cancel the h's:
{{{lim(h->0, 5)}}}
And since the limit of a constant is that constant, this simplifies to just
5<br>
So {{{lim(h->0, ((5x+5h+1)-(5x+1))/h) = 5}}}