By applying #lim_(x -> 1)f(x)#, the answer to #lim_(x -> 1)2x^2# is simply 2.
The limit definition states that as x approaches some number, the values are getting closer to the number. In this case, you can mathematically declare that #2(->1)^2#, where the arrow indicates that it approaches x = 1. Since this is similar to an exact function like #f(1)#, we can say that it must approach #(1,2)#.
However, if you have a function like #lim_(x->1)1/(1-x)#, then this statement has no solution. In hyperbola functions, depending on where x approaches, the denominator may equal zero, thus no limit at that point such exists.
To prove this, we can use #lim_(x->1^+)f(x)# and #lim_(x->1^-)f(x)#. For #f(x) = 1/(1-x)#,
#lim_(x->1^+)1/(1-x) = 1/(1-(x>1->1))=1/(-->0)=-oo#, and
#lim_(x->1^-)1/(1-x) = 1/(1-(x<1->1))=1/(+ ->0) = +oo#
These equations state that as x approaches to 1 from the right of the curve (#1^+#), it keeps going down infinitely, and as x approaches from the left of the curve (#1^-#), it keeps going up infinitely. Since these two parts of x = 1 do not equal, we conclude that #lim_(x->1)1/(1-x)# does not exist.
Here is a graphical representation:
graph{1/(1-x) [-10, 10, -5, 5]}
Overall, when it comes to limits, make sure to watch for any equation that has a zero in the denominator (including others like #lim_(x->0)ln(x)#, which does not exist). Otherwise you will have to specify if it approaches zero, infinity, or -infinity using the notations above. If a function is similar to #2x^2#, then you can solve for it by substituting x into the function using the limit definition.
Whew! It sure is a lot, but all the details are very important to note for other functions. Hope this helps!
