Existence of Limit: a and b Value?

[hehehe]

what value of the constants a and b if the following limit exists
lim (ax + |x + 1|)|x + b − 2| |x + 1|
x→−1

|x|= x for x≥ 0 and |x|= -x for x<0
|x+1|= x+1 for x≥ -1 and |x+1|= -(x+1) for x<-1
I don't know how to determine |x + b − 2| is positive or negative.

i know that if limit exists, lim x→ −1 ^- f(x) =lim x→ −1 ^+ f(x)

but i cannot cancell the factor x + 1

 

[hehehe]

Homework Statement

http://holland.pk/nwhxy2ji
2. Homework Equations

The Attempt at a Solution

|x|= x for x≥ 0 and |x|= -x for x<0
|x+1|= x+1 for x≥ -1 and |x+1|= -(x+1) for x<-1
I don't know how to determine |x + b − 2| is positive or negative.

i know that if limit exists, lim x→ −1 ^- f(x) =lim x→ −1 ^+ f(x)

i tired to solve it,but i cannot cancell the factor x + 1

 

[Samy_A]

Homework Statement

http://holland.pk/nwhxy2ji
2. Homework Equations [/B]

The Attempt at a Solution

|x|= x for x≥ 0 and |x|= -x for x<0
|x+1|= x+1 for x≥ -1 and |x+1|= -(x+1) for x<-1
I don't know how to determine |x + b − 2| is positive or negative.

i know that if limit exists, lim x→ −1 ^- f(x) =lim x→ −1 ^+ f(x)

i tired to solve it,but i cannot cancell the factor x + 1

I've not solved the whole exercise, but you could start by simplifying the problem somewhat.
Clearly the denominator is 0 when x=-1. So if the numerator isn't equal to 0 in x=-1, there is no way the limit will exist.
What can you tell about a and/or b if you require the numerator to be 0 in x=-1?

 

[hehehe]

I've not solved the whole exercise, but you could start by simplifying the problem somewhat.
Clearly the denominator is 0 when x=-1. So if the numerator isn't equal to 0 in x=-1, there is no way the limit will exist.
What can you tell about a and/or b if you require the numerator to be 0 in x=-1?

(ax+ (x+1)) |x+b-2| =0 for x≥-1
(ax+ (x+1))=0
a=-(x+1)/x

or |x+b-2| =0
b=2-x

(ax-(x+1)) |x+b-2| =0 for x<-1
a=(x+1)/x

or |x+b-2| =0
b=2-x

(x+1)/x =-(x+1)/x
x=-1

b=2-(-1)=3

a=-1 and b=3 when limit will exist
is it right ?
thank you

 

[Samy_A]

(ax+ (x+1)) |x+b-2| =0 for x≥-1
(ax+ (x+1))=0
a=-(x+1)/x

or |x+b-2| =0
b=2-x

(ax-(x+1)) |x+b-2| =0 for x<-1
a=(x+1)/x

or |x+b-2| =0
b=2-x

(x+1)/x =-(x+1)/x
x=-1

b=2-(-1)=3

a=-1 and b=3 when limit will exist
is it right ?
thank you

No, you made an error somewhere.

You make it way too complicated. You don't (at this stage) have to distinguish between x≤-1 and x≥-1.
All you have to do is set the numerator at x=-1 to 0.
That will give you necessary conditions on a and/or b.

After that, you still will have to check if the limit exists if a and/or b meet the conditions, and if the limits exists, compute it. This all turns out to be rather straightforward, though.

 

[hehehe]

No, you made an error somewhere.

You make it way too complicated. You don't (at this stage) have to distinguish between x≤-1 and x≥-1.
All you have to do is set the numerator at x=-1 to 0.
That will give you necessary conditions on a and/or b.

After that, you still will have to check if the limit exists if a and/or b meet the conditions, and if the limits exists, compute it. This all turns out to be rather straightforward, though.

at x=-1
(ax+ (x+1)) |x+b-2| =0
-a|b-3| =0
a=0 , b=3 but i have a problem , can |b-3| = b-3?
i put a=0 and b=3 to the lim
and do it in left-hand limits and right-hand limits

left-hand limits=right-hand limits=0 ,so it exists

is it right?

Samy_A thank you

 

[Samy_A]

at x=-1
(ax+ (x+1)) |x+b-2| =0
-a|b-3| =0
a=0 , b=3 but i have a problem , can |b-3| = b-3?

Yes, this is correct. At least if you meant that a=0 or b=3. These two conditions don't have both to be true (although they can both be true of course).

Whether b-3 is positive or negative in the case a=0 doesn't matter: 0|b-3| will be 0 anyway.

i put a=0 and b=3 to the lim
and do it in left-hand limits and right-hand limits

left-hand limits=right-hand limits=0 ,so it exists

is it right?

I'm not sure I understand what you did here. The limit isn't necessarily 0 though.
I would treat the two cases (a=0, b=3) separately.

For example, in the case a=0, the expression becomes:$\frac{|x+1||x+b-2|}{|x+1|}=|x+b-2|$ for $x \neq -1$. Calculating the limit for $x \to-1$ is then straightforward.
Similarly for the case b=3.

Samy_A thank you

You are welcome.

 

[hehehe]

Whether b-3 is positive or negative in the case a=0 doesn't matter: 0|b-3| will be 0 anyway.
I'm not sure I understand what you did here. The limit isn't necessarily 0 though.
I would treat the two cases (a=0, b=3) separately.

For example, in the case a=0, the expression becomes:$\frac{|x+1||x+b-2|}{|x+1|}=|x+b-2|$ for $x \neq -1$. Calculating the limit for $x \to-1$ is then straightforward.
Similarly for the case b=3.

$|x+b-2|$ for $x \neq -1$ , i don't know what is your meaning
if i just substitute -1 to the limit , then i get lim $x \to-1$ $|x+b-2|$ the next step is that do it in left-hand limits and right-hand limits?
but i don't not have to distinguish |x+b-2|?

 

[Samy_A]

$|x+b-2|$ for $x \neq -1$ , i don't know what is your meaning
if i just substitute -1 to the limit , then i get lim $x \to-1$ $|x+b-2|$ the next step is that do it in left-hand limits and right-hand limits?
but i don't not have to distinguish |x+b-2|?

Let's give the expression a name: $\displaystyle f(x)=\frac{(ax+|x+1|)|x+b-2|}{|x+1|}$ (for $x \neq -1)$.
You are asked to establish whether $\displaystyle \lim_{x\rightarrow -1} f(x)$ exists, and if so, what the limit is.

We already know that the limit can only exist if a=0 or b=3 (or both).

For a=0, we can simplify the expression for $f(x)$:
$\displaystyle f(x)=\frac{(ax+|x+1|)|x+b-2|}{|x+1|}=\frac{(0x+|x+1|)|x+b-2|}{|x+1|}=\frac{|x+1||x+b-2|}{|x+1|}=|x+b-2|$ (for $x \neq -1$).

So, in the case a=0, you have to compute $\displaystyle \lim_{x\rightarrow -1} f(x)=\lim_{x\rightarrow -1}|x+b-2|$.
That's an easy limit to compute, no need to distinguish left-hand and right-hand limits.

And similarly for the other case, b=3.

 

[hehehe]

Let's give the expression a name: $\displaystyle f(x)=\frac{(ax+|x+1|)|x+b-2|}{|x+1|}$ (for $x \neq -1)$.
You are asked to establish whether $\displaystyle \lim_{x\rightarrow -1} f(x)$ exists, and if so, what the limit is.

We already know that the limit can only exist if a=0 or b=3 (or both).

For a=0, we can simplify the expression for $f(x)$:
$\displaystyle f(x)=\frac{(ax+|x+1|)|x+b-2|}{|x+1|}=\frac{(0x+|x+1|)|x+b-2|}{|x+1|}=\frac{|x+1||x+b-2|}{|x+1|}=|x+b-2|$ (for $x \neq -1$).

So, in the case a=0, you have to compute $\displaystyle \lim_{x\rightarrow -1} f(x)=\lim_{x\rightarrow -1}|x+b-2|$.
That's an easy limit to compute, no need to distinguish left-hand and right-hand limits.

And similarly for the other case, b=3.

$\lim_{x\rightarrow -1}|x+b-2| = |b-3|$ but i don't know that value of b when a=0 ,how can i compute.
similarly,when b=3, $\lim_{x\rightarrow -1}(ax+|x+1|) = -a$ ,it is same with the above one
when a=0 ,b=3 ,$\displaystyle \lim_{x\rightarrow -1} f(x) = 0$

 

[Samy_A]

$\lim_{x\rightarrow -1}|x+b-2| = |b-3|$ but i don't know that value of b when a=0 ,how can i compute.
similarly,when b=3, $\lim_{x\rightarrow -1}(ax+|x+1|) = -a$ ,it is same with the above one
when a=0 ,b=3 ,$\displaystyle \lim_{x\rightarrow -1} f(x) = 0$

Correct.
When a=0, the limit is |b-3|. You can't say more. This limit exists for all values of b.
Similarly, when b=3, the limit is -a. This limit exists for all values of a.

 

[hehehe]

Correct.
When a=0, the limit is |b-3|. You can't say more. This limit exists for all values of b.
Similarly, when b=3, the limit is -a. This limit exists for all values of a.

thank you, you are a patient and good teacher.
you have a full explanation ,thank you again.