Limits are fundamental for both differential and integral calculus. The formal definition of a derivative involves a limit as does the definition of a definite integral.
The limit of a function(if it exists) for some x-value, a is the height the function gets closer and closer to as x gets closer and closer to a from the left and the right. Lets take a look at what that even means.
The limit of the function f(x) = 3x + 1 as x approaches 2. When writing a limit we denote this as the following equation.
lim f(x) = lim 3x+1.
Now for such an easy limit we can plug in the number 2 and solve this limit. Plugging in the 2 we see.
lim f(x) = lim 3(2)+1 = 7
That the limit is 7. So as x approaches 2 from the left and right side. The limit of this function is 7.
Now your probably think well that's silly. Any function that's continuous you just plug in and solve for x. Now limits are important for functions with holes.
Definition of a Limit:
lim f(x) exists if and only if
x -> a
1. lim f(x) exists
x->a- (a from the left side)
2. lim f(x) exists, and
x->a+(a from the right side)
3. lim f(x) = lim f(x)
x -> a- x->a+
Number three is the most important.
A function with an infinite limit and vertical asymptotes.
f(x) = (x+2)(x-5) / (x-3)(x+1) This function has a VA at x = 3, x = -1
So the lim f(x) = lim (3+2)(3-5) / (3-3)(3+1) = infinity as x will never reach 0 thus it heads closer to 0
x-> 3- x->3-
From the right side the limit will be negative infinity. Thus we know by the limit definition that the limit does not exist for x approaches 3.
Definition of Continuity:
A function is continuous at a point x = a if the following three conditions are satsified.
1.f(a) is defined
2.limf(x) exists and,
3. f(a) = lim f(x)
Limits to Memorize:
lim c = c
lim 1/x = infinity
lim 1/x = negative infinity
lim 1/x = 0
x -> infinity(same for both - and + infinity)
lim sin x/x = 1
lim cos x - 1/x =0
lim (1+1/x)^x = e
x -> infinity
Plugging and Chugging Limits:
Any limit where it's continuous and straight forward.
lim (x^2 - 10) = -1
lim 10/x-5= 10/0 if you get an answer that has any numerator thats not zero divided by zero. The limit
x - > 5
does not exist.
Now in the below case if you get 0/0 you have a real limit problem.
lim (x^2 - 25) / (x - 5) = 0/0 which is undefined
undefined is not the limit. This is not the answer we are looking for. Four things we can do to find a limit that is undefined. Calculator, algebra, limit sandwhich, L'Hopitals rule.
I will only show the algebra and L'Hopitals rule. I will not explain L'Hopitals Rule though.
Using algebra to solve for
lim (x^2-25)/(x-5) we can rewrite this by factoring (x^2 - 5) = (x-5)(x+5) knowing this we can
divide out the (x-5) which leaves us with the equation lim x+5 = 10
Using L'Hoptials Rule
lim (x^2-25)/(x-5) = lim d/dx(x^2 - 25) / d/dx(x-5) = lim 2x = 10.
x-> 5 x->5 x->5