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!
