Calculus

>>204816
Check the type (insert x=2 in nominator/denominator) : 0/0

Now you can use the rule of L'Hospital or you can try to eliminate the factor (x-2) in both nominator and denominator.

>>204833
Isn't is easier to just derive the top and bottom?

>>204883
That is L'hopitals rule.

>>204833
What this anon said, but i'd recommend the second solution if you're still in highschool (I don't remember learning l'Hôpital's rule in highschool, i could be wrong...)

>>204816
x^3 - 7x + 6 = (x-2)*(x-1)*(x+3)

3x^2 - 8x + 4 = 3*(x-2)*(x-2/3)

Simplifying by (x-2) gives us that the limit from the picture is the same as

lim (x->2) [(x-1)*(x+3)]/[3*(x-2/3)] which is 5/4

You're welcome