Jimmy84 said:
Homework Statement
Im teaching myself calculus and I am trying to solve 40 problems about limits for now I just have been having troubles with the following problems.
-First part
Proff the limit of
lim x² - 1/ x + 1 = -2
x --> -1
-Second part
Determine the limit of the following problems
1.)
lim (z³ + 8)
z --> -2
2.)
lim cubit root of x² -3x +4 / 2x² -x -1
x --> 4
3.)
lim 3 - square root of x/ 9 -x
x --> 9
4.)
lim y³ + 8 / y + 2
y --> -2
5.)
lim square root of (x ) -1 /x -1
x --> 1
Homework Equations
The Attempt at a Solution
-First part
Proff the limit of
lim x² - 1/ x + 1 = -2
x --> -1
0 < |x +1| < delta then |(x² - 1/ x + 1) +2|< epsilon
|(x-1) (x+1) / (x+1) +2 | epsilon
after simplifying
|(x-1) +2 | epsilon
|x + 1| epsilon
Then I am confused about what to do afterwards since I can't multiply epsilon by a constant to get delta.
Im also having troubles in order to factorize the problems in part 2, I have no clue about how to start, I would appreaciate some help, thanks a lot.
You are almost done. The thing you "multiply" by is 1. Just remember what it is you are trying to do! To prove "\lim_{x\to a} f(x)= L", you have to prove ""Given any \epsilon> 0, there exist \delta> 0 such that if |x-a|< \delta, |f(x)-L|< \epsilon".
Here |f(x)- L|< \epsilon is |(x^2- 1)/(x+1)|< \epsilon and |x-a|< \delta is |x-(-1)|= |x+1|<\delta. Now you have succesfully reduced |(x^2- 1)/(x+1)|< \epsilon to |x+1|< \epsilon. Do you see that you can now just take \delta= \epsilon? The crucial point is that if you take \delta= \epsilon you can say if |x- (-1)|< \delta then |x+1|< \epsilon and then
reverse what you have done, arriving at |(x^2- 1)/(x+1)|< \epsilon, exactly what you wanted. (Normally, you don't actually do that reversal. Once you have shown how to find \delta from \epsilon, as long as everything you did
is reversible, you know you will get the right result.)
For the others, where you are only asked to
find the limit, not use the "\epsilon-\delta" definition to prove it,
1.)
lim (z³ + 8)
z --> -2
This is easy- just use the basic laws of limits: lim fg= (lim f)(lim g), lim f+g= lim f+ lim g, \lim_{x\rightarrow a} x= a, and lim c= c when c is a constant, to show that this limit is just (-2)^2+ 8
2.)
lim cubit root of x² -3x +4 / 2x² -x -1
x --> 4
The denominator is 2(4)^2- 4- 1= 32- 4-1= 27 which is NOT 0! It is only when a denominator is 0 that you have to factor and try to cancel to get rid of that. It's unclear whether your cube root is of the whole fraction or only the numerator (use parentheses!). That will affect the result but the point about the denominator is valid either way.
3.)
lim 3 - square root of x/ 9 -x
x --> 9
Again, please use parentheses! Do you mean (3- sqrt(x))/(9- x)? If so use "a^2- b^2= (a-b)(a+b) with a= 3 and b= \sqrt{3}.
4.)
lim y³ + 8 / y + 2
y --> -2
I presume you know "a^2- b^2= (a-b)(a+b)". It is also true that a^3- b^3= (a- b)(a^2+ ab+ b^2 and a^3+ b^3= (a+ b)(a- ab+ b). In fact, for any integer n, a^n- b^n= (a-b)(a^{n-1}+ a^{n-2}b+ \cdot\cdot\cdot+ ab^{n-2}+ b^{n-1} and, for any
odd integer n, a^n+ b^n= (a+b)(a^{n-1}- a^{n-2}b+ \cdot\cdot\cdot- ab^{n-2}+ b^{n-1}).
But you don't really NEED to know those. You should think: "Clearly, the denominator, y+ 2, goes to 0 as y goes to -2 because (-2)+ 2= 0. If the numerator did NOT go to 0, there would be NO limit (the fraction would "go to infinity"). Checking, sure enough (-2)^3+ 8= -8+ 8= 0. But just like you can solve "x^3+ 8= 0" factoring, knowing that -2 makes that 0 tell us that (x-(-2))= (x+ 2) is a factor of x^3+ 8. We can find the other factor by dividing x^3+ 8 by x+ 2.
5.)
lim square root of (x ) -1 /x -1
x --> 1
Again, this is an application of a^2- b^2= (a-b)(a+b) with a= \sqrt{x} and b= 1.
