I already know how to compute a limit by its definition when x tends to a real number. But know I questioned myself how can you prove a limit if it tends to infitity?. I looked for an exercise and found this:
$$\lim_{x\to\infty} = \frac{7x+2}{4x+3}-\frac{7}{4}$$
I tried to follow the normal path but got stucked in here:
$|\frac{13}{4(4x+3)}|<\epsilon ; |x-\infty|<\delta$
UPDATE: I kept operating and came up with: As $x\to+\infty;|\frac{13}{4(4x+3)}|=\frac{13}{4(4x+3)}$
And if $\frac{13}{4(4x+3)}<\epsilon\to\frac{4(4x+3)}{13}>\frac{1}{\epsilon}$
I think I'm near to the solution, but I fail on seeing what is the relation between $k$ and $\epsilon$
I would appreciate any help!
HINT:
Note that
$$\left|\frac{7x+2}{4x+3}-\frac74\right|=\left|\frac{13}{4(4x+3)}\right|$$
Now show that for any $\epsilon>0$, there is a number $B>0$ such that $\left|\frac{13}{4(4x+3)}\right|<\epsilon$ whenever $x>B$.