# Homework Help: Limit Problems

1. Jun 7, 2009

### Jimmy84

1. The problem statement, all variables and given/known data

Im teaching myself calculus and im 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

2. Relevant equations

3. 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 im confused about what to do afterwards since I cant 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.

2. Jun 7, 2009

### Bohrok

In proving the first limit, why not choose δ = ε ? Seems that this one works out well where you don't need to multiply or divide epsilon by a number when choosing delta.

For 1 you can evaluate the limit with substitution since it won't result in division by zero.

or look it up on Google (I've never used the box method myself). For number 3 and 5, look up multiplying by the conjugate for getting rid of square roots like the ones in those. For 4 look up factoring sum of cubes where you have x3.

3. Jun 7, 2009

### HallsofIvy

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,
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 $(-2)^2+ 8$

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.

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 $b= \sqrt{3}$.

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.

Again, this is an application of $a^2- b^2= (a-b)(a+b)$ with $a= \sqrt{x}$ and b= 1.