Limits Homework Help

[ambright4life]

I need help with 2 Calculus homework problems that I can't figure out for nothing.

1. Find lim x³+1/x+1 graphically as it approaches -1

2. Find the limit L. Then find delta (symbol) > 0 such that |f(x)-L| < 0.001 whenever 0 < |x-c| < delta (symbol).

lim (5x+2) as it approaches 2

For #1 I know the equation I will get will be x²-x+1 and after that I get confused.

For #2 I don't know what to do at all!

So pleaseeeee help! I will greatly appreciated!

 

[MarkFL]

1.) I assume you mean:

\(\displaystyle \displaystyle \lim_{x\to-1}\frac{x^3+1}{x+1}\)

You've probably noticed when you substitute -1 in for x you get the indeterminate form 0/0. Unless you have been taught L'Hôpital's rule, you are going to need to use some algebra to rewrite the expression to get rid of the zero in the denominator. Is there any way to factor the numerator, where \(\displaystyle x+1\) is a factor?

Yes, you've noticed the numerator will factor as the sum of cubes:

\(\displaystyle \displaystyle \lim_{x\to-1}\frac{x^3+1}{x+1}=\lim_{x\to-1}\frac{(x+1)(x^2-x+1)}{x+1}=\lim_{x\to-1}(x^2-x+1)\)

Now, try substitution again.

2.) First, what is \(\displaystyle L\)?

 

[DrPhil]

1. Find lim x³+1/x+1 graphically as x approaches -1

"graphically" suggests you should plot the parabola .. since it is a continuous function, you can read off the value when x=-1.

2. Find the limit L. Then find delta (symbol) > 0 such that |f(x)-L| < 0.001 whenever 0 < |x-c| < delta (symbol).

lim (5x+2) as x approaches 2

The symbol "L" represents the limit, which you can find by substituting x=2 in the expression.

Then the problem looks asks how close x has to be to 2 in order that the value of the expression is within 0.001 of the limit. What is the value of delta such that if x is closer to 2 than delta, then f(x) is guaranteed to be within 0.001 of L.

 

[Deleted member 4993]

I need help with 2 Calculus homework problems that I can't figure out for nothing.

1. Find lim (x³+1)/(x+1) graphically as it approaches -1→ Need to use grouping symbols as shown to indicate proper order of operations.

2. Find the limit L. Then find delta (symbol) > 0 such that |f(x)-L| < 0.001 whenever 0 < |x-c| < delta (symbol).

lim (5x+2) as it approaches 2

For #1 I know the equation I will get will be x²-x+1 and after that I get confused.

For #2 I don't know what to do at all!

So pleaseeeee help! I will greatly appreciated!

.

 

[ambright4life]

Thank you all for the help. Okay this is what I got for my answers.

#1. The limit is 1.
As x --> -1 from the left f(x) ---> 1
As x --> -1 from the right f(x) ---> 1

#2. Since 0 < |x + 14/5| < delta (symbol) and |x + 14/5| < 0.0002 then delta (symbol) = 0.0002

is this correct?

 

[JeffM]

Thank you all for the help. Okay this is what I got for my answers.

#1. The limit is 1.
As x --> -1 from the left f(x) ---> 1
As x --> -1 from the right f(x) ---> 1

#2. Since 0 < |x + 14/5| < delta (symbol) and |x + 14/5| < 0.0002 then delta (symbol) = 0.0002

is this correct?

Here is how you can check your own work on the next test.

\(\displaystyle \displaystyle \left(\lim_{x \rightarrow 2}5x + 2\right) = \left(\lim_{x \rightarrow 2}5x\right) + \left(\lim_{x \rightarrow2}2\right) = 5\left(\lim_{x \rightarrow 2}x\right) + 2 = 5(2) + 2 = 12.\)

\(\displaystyle 0 < |x - 2| < 0.0002 = \delta \implies -0.0002 < x - 2 < 0.0002 \implies 1.9998 < x < 2.0002 \implies 9.999 < 5x < 10.001 \implies\)

\(\displaystyle 11.999 < (5x + 2) < 12.001 \implies 11.999 - 12 < {(5x + 2) - 12} < 12.001 -12 \implies |(5x + 2) - 12| < 0.001.\)

So \(\displaystyle \delta = 0.0002\) is correct.

 

[ambright4life]

Thank you JeffM! Is #1 right too?

 

[DrPhil]

1. Find lim (x³+1)/(x+1) graphically as it approaches -1

For #1 I know the equation I will get will be x² - x + 1 and after that I get confused.

That statement of your original work was OK. However your final result is not correct.

Look again at your calculation of the limit. Be sure you are evaluating near x = minus 1.

 

[ambright4life]

So the limit is not 1? I don't know what I'm doing wrong.

 

[MarkFL]

You are substituting 1 for x, rather than -1. You want:

\(\displaystyle (-1)^2-(-1)+1=?\)

 

[ambright4life]

So the limit is 3 right?

 

[MarkFL]

Yes, that's correct. :cool: