Simple Limit Problems *I need ur help*

[heartnet]

Hi there,
I have few limits question to ask because I am not understanding it..Just basic one..but Im too 'slow' I guess
I will post few of it first in this first post
1)Absolute value related
lim -5|x+5|/x+5
x>-5

the answer is limit d.n.e..Im confused :?
2)About pice-wise function
lim f(x) -2x+3 x doesnt equal to -2
x>2::::::-5:::::x = 2

lim f(x) 3x+1 x doesnt equal to 3
x>3::::::-5:::::x = 3

Question 3 is uploaded as attachment
Please point my fault with explanations..the answer should be -1/3

 

[tkhunny]

For x >= -5, |x+5| = (x+5)

For x < -5, |x+5| = -(x+5)

Do both cases separately and think about the concept of equality.

 

[lookagain]

1) Absolute value related

\(\displaystyle \lim_{x \to \ -5} \frac{ -5|x+5|}{x+5}\)


Let \(\displaystyle u = x + 5.\)

When \(\displaystyle x = -5, \ u = 0.\)

The limit becomes

\(\displaystyle \lim_{u \to 0} \frac{|u|}{u}\)


Check the left-hand and right-hand limits:

\(\displaystyle \lim_{u \to 0^{-}} \frac{|u|}{u} = -1\)

\(\displaystyle \lim_{u \to 0^{+}} \frac{|u|}{u} = 1\)


Because the limits from the left side of \(\displaystyle 0\) and the right side of \(\displaystyle 0\) are not equal
to the same value, then this limit does not exist.

The answer is "The limit does not exist."

The above contains my edit of your problem. Read it and see if you are no longer confused
about this first problem. Yes, and in general for all of your problems, (continue to) show some
attempted work on all of your problems.

 

[BigGlenntheHeavy]

\(\displaystyle f(x) \ = \ \lim_{x\to-5}\frac{5|x+5|}{(x+5)}, \ the \ Marqui \ can't \ help \ us \ on \ this \ one.\)

\(\displaystyle |x+5| \ = \ x+5 \ if \ x+5 \ \ge \ 0 \ \implies \ x \ \ge \ -5,\)

\(\displaystyle or \ |x+5| \ = \ -(x+5) \ if \ x+5 \ < \ 0 \ \implies \ x \ < \ -5.\)

\(\displaystyle Ergo, \ we \ have \ \lim_{x\to-5^+}\frac{5(x+5)}{(x+5)} \ = \ 5\)

\(\displaystyle but, \ we \ also \ have \ \lim_{x\to-5^-}\frac{-5(x+5)}{(x+5)} \ = \ -5\)

\(\displaystyle Hence, \ since \ we \ have \ two \ different \ limits, \ the \ limit \ of \ f(x) \ doesn't \ exist.\)

\(\displaystyle See \ graph.\)

[attachment=0:4o1cqxag]aaa.jpg[/attachment:4o1cqxag]

 

[heartnet]

thanks lookagain and BigGlenntheHeavy
what about the 2nd question?
Actually I wish to show my work but I don't know how to begin on this piece-wise function

 

[tkhunny]

1) That really hurts.
2) It is the same. Find the one-sided limits and see that/if they are the same.

 

[heartnet]

1) That really hurts.
2) It is the same. Find the one-sided limits and see that/if they are the same.

1) what do you mean by 'hurts'?
2)I know..but what that -5 for? I mean the limit is -5 as x>3? or should I use the above one..please explain

 

[BigGlenntheHeavy]

\(\displaystyle \lim_{x\to-\infty}\frac{\sqrt{x^2+2}}{(3x-6)} \ = \ \lim_{x\to-\infty}\frac{\sqrt {x^2}}{3x} \ = \ \lim_{x\to-\infty}\frac{|x|}{3x} \ = \ -\frac{1}{3}.\)

\(\displaystyle Note: \ Infinity \ is \ not \ a \ number.\)

\(\displaystyle See \ graph, \ note \ x \ = \ 2 \ is \ a \ vertical \ asymptote \ and \ y \ = \ \pm\ \ \frac{1}{3} \ are \ horizontal \ asymptotes.\)

[attachment=0:2c267ck7]bbb.jpg[/attachment:2c267ck7]

 

[lookagain]

\(\displaystyle >> \lim_{x\to-\infty}\frac{\sqrt x^2}{3x} \ << = \ \lim_{x\to-\infty}\frac{|x|}{3x} \ = \ -\frac{1}{3}.\)

heartnet,

\(\displaystyle this \ should \ be \ \ \lim_{x\to\ -\infty}\frac{{\sqrt{x^2}}}{3x}}, \ as \ \sqrt{x}^2 = x,\ which \ would \ not \ be \ what \ you \ intend \ \ in \ the \ numerator.\)

\(\displaystyle Instead,\ in \ the \ numerator, \ we \ need \ \sqrt{x^2} = |x|.\)

 

[heartnet]

\(\displaystyle >> \lim_{x\to-\infty}\frac{\sqrt x^2}{3x} \ << = \ \lim_{x\to-\infty}\frac{|x|}{3x} \ = \ -\frac{1}{3}.\)

heartnet,

\(\displaystyle this \ should \ be \ \ \lim_{x\to\ -\infty}\frac{{\sqrt{x^2}}}{3x}}, \ as \ \sqrt{x}^2 = x,\ which \ would \ not \ be \ what \ you \ intend \ \ in \ the \ numerator.\)

\(\displaystyle Instead,\ in \ the \ numerator, \ we \ need \ \sqrt{x^2} = |x|.\)

ok thanks lookagain and BigGlenntheHeavy
what about my 2nd question

 

[heartnet]

anyone? my 2nd question

 

[lookagain]

\(\displaystyle Sorry \ lookagain, \ I \ re-edited \ to \ keep \ you \ happy.\)

It's *not* to keep "me happy," it's to be correct.

 

[Deleted member 4993]

Hi there,
2)About pice-wise function
lim f(x) -2x+3 x doesnt equal to -2
x>2::::::-5:::::x = 2

lim f(x) 3x+1 x doesnt equal to 3
x>3::::::-5:::::x = 3

Question 3 is uploaded as attachment
Please point my fault with explanations..the answer should be -1/3

Please show us your work, indicating exactly where you are stuck - so that we may know where to begin to help you.

 

[heartnet]

Hi there,
2)About pice-wise function
lim f(x) -2x+3 x doesnt equal to -2
x>2::::::-5:::::x = 2

lim f(x) 3x+1 x doesnt equal to 3
x>3::::::-5:::::x = 3

Question 3 is uploaded as attachment
Please point my fault with explanations..the answer should be -1/3

Please show us your work, indicating exactly where you are stuck - so that we may know where to begin to help you.

the thing is I dont know where to begin
2)
a)clueless
b)I know that we must see limit as x approach 3 from right and left..so let say I choose the first 3x+1 so it will be 7 for both..err so limit is 7? isnt the 2nd eq show -5? im confuse about piece wise..please..I wish I could show my works for both 2a and 2b but I dont know where to begin