Finding Value of a & b for Limit Existence

[steven cheung]

Homework Statement

For what value of the constants a and b such that the following limit exists?

lim {(ax+|x+1|)|x+b-2|}/|x+1|
x->-1 help me ,thx

Homework Equations

The Attempt at a Solution

first, I know that I should cancel the absolute value at denominator of x+1. but i don't how to do. also x+1 if x>1. -(x+1) if x<1
[/B]

 

[Samy_A]

Homework Statement

For what value of the constants a and b such that the following limit exists?

lim {(ax+|x+1|)|x+b-2|}/|x+1|
x->-1 help me ,thx

Homework Equations

The Attempt at a Solution

Hello Steven,

You have to show an "attempt at a solution" before we are allowed to help.

To help you start: what is the first thing you try when you have to evaluate such a limit? And why doesn't it work in this case?

 

[Samy_A]

The Attempt at a Solution

first, I know that I should cancel the absolute value at denominator of x+1. but i don't how to do. also x+1 if x>1. -(x+1) if x<1[/B]

What value must the expression in the numerator have in x=-1 in order for the limit for x → -1 to possibly exist?

 

[steven cheung]

What value must the expression in the numerator have in x=-1 in order for the limit for x → -1 to possibly exist?

i do not understand
have some tip?

 

[Samy_A]

i do not understand
have some tip?

Say you have to establish whether the following limit exists:
$\displaystyle \lim_{x\rightarrow -1} \frac{x+7}{|x+1|}$.
How would you do that, and what is the answer?

Then look at the limit you have to compute:
$\displaystyle \lim_{x\rightarrow -1} \frac{(ax+|x+1|)|x+b-2|}{|x+1|}$
What value must the numerator $(ax+|x+1|)|x+b-2|$ have for x=-1 in order for this limit to possibly exist?

 

[steven cheung]

Say you have to establish whether the following limit exists:
$\displaystyle \lim_{x\rightarrow -1} \frac{x+7}{|x+1|}$.
How would you do that, and what is the answer?

 

[Samy_A]

I don't understand how you get the 2 or -2 in the denominators as $|1+(-1)|=0$.

Just plugging in $x=-1$ in $\frac{x+7}{|x+1|}$ gives you $\frac{6}{0}$, obviously not a very nice limit.
See how the problem is not only that the denominator is 0 in $x=-1$, but also that the numerator is 6. If the numerator was also 0, then maybe (just maybe), the limit could exist.

That should give you a huge clue about what should happen with the numerator $(ax+|x+1|)|x+b-2|$ at $x=-1$ for the limit to maybe (just maybe) exist.

 

[steven cheung]

I don't understand how you get the 2 or -2 in the denominators as $|1+(-1)|=0$.

Just plugging in $x=-1$ in $\frac{x+7}{|x+1|}$ gives you $\frac{6}{0}$, obviously not a very nice limit.
See how the problem is not only that the denominator is 0 in $x=-1$, but also that the numerator is 6. If the numerator was also 0, then maybe (just maybe), the limit could exist.

That should give you a huge clue about what should happen with the numerator $(ax+|x+1|)|x+b-2|$ at $x=-1$ for the limit to maybe (just maybe) exist.

(-a)(b-3)>0
b-3>0 or -a>0
b>3 0r a <0
that right?

 

[Samy_A]

(-a)(b-3)>0
b-3>0 or -a>0
b>3 0r a <0
that right?

(Please always explain how you get a result, else it may be difficult to help you along.)

I assume you got $(-a)(b-3)$ by setting $x=-1$ in the numerator $(ax+|x+1|)|x+b-2|$.
That actually gives $-a|b-3|$. Why do you want that numerator to be larger than 0? How does >0 help you to establish that a limit may exist?

 

[steven cheung]

(-a)(b-3)>0
b-3>0 or -a>0
b>3 0r a <0
that right?

(Please always explain how you get a result, else it may be difficult to help you along.)

I assume you got $(-a)(b-3)$ by setting $x=-1$ in the denominator $(ax+|x+1|)|x+b-2|$.
That actually gives $-a|b-3|$. Why do you want that denominator to be larger then 0? How does >0 help you to establish that a limit may exist?

so, -a|b-3| if put in x=-1
then, a is all real number, b is 3 ?

 

[Samy_A]

so, -a|b-3| if put in x=-1
then, a is all real number, b is real munber(except 3) ?

I have asked the key question a number of times: what value must the numerator have in $x=-1$ in order for the limit to possibly exist?
You now know that the numerator is $-a|b-3|$ when $x=-1$.
If the numerator is, say, equal to 7, you get $\frac{7}{0}$ when $x=-1$. Can the limit then exist?

What is the only value for the numerator $-a|b-3|$ that may makes it possible for a limit to exist for $x\to -1$?

 

[steven cheung]

I have asked the key question a number of times: what value must the numerator have in $x=-1$ in order for the limit to possibly exist?
You now know that the numerator is $-a|b-3|$ when $x=-1$.
If the numerator is, say, equal to 7, you get $\frac{7}{0}$ when $x=-1$. Can the limit then exist?

What is the only value for the numerator $-a|b-3|$ that may makes it possible for a limit to exist for $x\to -1$?

i need some time to think,because i learn limit on 3 days

 

[Samy_A]

i need some time to think,because i learn limit on 3 days

The more theoretical point is as follows.
Say you have two continuous functions $f:\mathbb R \to \mathbb R$ and $g:\mathbb R \to \mathbb R$.

Does $\displaystyle \lim_{x\rightarrow -1} \frac{f(x)}{g(x)}$ exist, and if so, what is the limit?

If $g(-1) \neq 0$, the answer is easy: $\displaystyle \lim_{x\rightarrow -1} \frac{f(x)}{g(x)}=\frac{f(-1)}{g(-1)}$.

It get's more complicated if $g(-1) = 0$. Then just plugging in $x=-1$ would give $\displaystyle \lim_{x\rightarrow -1} \frac{f(x)}{g(x)}=\frac{f(-1)}{0}$.
Now dividing a real number by 0 is not possible, so if, for example, $f(-1)=7$, we can conclude that $\displaystyle \lim_{x\rightarrow -1} \frac{f(x)}{g(x)}$ doesn't exist (or maybe is $\pm \infty$).
It is only when plugging in $x=-1$ gives $\frac{0}{0}$ (that's called an indeterminate form), that a finite limit may exist. Notice the may. It's not sure the limit will exist, but it is possible. It will depend on the precise behavior of $f$ and $g$ near $x=-1$.