What is the approach to solving limits using L'Hopital's rule?

[Jadenag]

Homework Statement

Homework Equations

The Attempt at a Solution

I need to make sure what I am doing is correct. My teacher hasnt been very helpful.
In the first two parts iwe merely placed in the limit value and gotten ratios which I think are the limits.

In part c and onwards iwe generally gotten an indeterminate form each time Iwe placed in the limit so i used L hopitals rule and differentiated the top and bottom functions and then placed in the limit value on i.e f'(a)/g'(a) and used that as my answer. Is my thought process correct?

The answers I achieved are
a 2
b 272/11
c 21.6
d -1/6
e 1/3
f 1
g 3

Some of these are scary fractions and decimals. Should I be worried about those. comments and criticism is welcome. Thanks in advance!

 

[QuarkCharmer]

c,e,f, and g are incorrect. g actually has 2 cases (or is undefined depending on how you look at it).

If you show your work we can figure out what happened.

 

[Jadenag]

C. i differentiated the top and the bottom. So I got
4x^3/(4x-5)
Next I put in the value 3 in place of x. Okay wait so I am getting 108/7 is this correct?

E.
Again I differentiate.
I get 1/(1/3x^-2/3) Put 27 in for x and I get 27. Is this the correct answer?

F.
Initially what I did was cancel out the x from all the terms.
And I am left with

(x^2 -7)/x^2 but i notice that that's not getting me any where

so i differentiated top and bottom from the beginning.
So I got (3x^2 -7)/3x^2

this wasnt getting me anywhere so I differentiated this again
and got 6x/6x. Here I am left with 1 hence my answer.

G.
I again did the same.
(2x+1)/1 and put in x as 1. to get 3.

Please let me know what i did wrong.

Also a question regarding L hopitals rule its NOt only applicable when x tends to zero right? Because Iwe used it in cases where x has not tended to zero as well.

 

[symbolipoint]

Item #c can be simplied and then direct substitution of x=3 can be done. No need for L'Hopital's Rule.

$Lim_{x\rightarrow 3}\frac{x^4-81}{2x^2-5x-3}$

Working with the expressin, $\frac{(x^2+9)(x^2-9)}{(2x+1)(x-3)}$
$\frac{(x^+9)(x+3)(x-3)}{(2x+2)(x-3)}$
$\frac{(x^2+9)(x+3)}{(2x+1)}$
Now you can substitute 3 for x in this last expression and evaluate.

 

[symbolipoint]

#f, L'Hopital's Rule gives these expressions,

First derivative of top and bottom:
$\frac{3x^2-7}{3x^2}$
Second derivatives...
$\frac{6x-0}{6x}$, which is clearly equal to 1 (that is, as x approaches 0).

 

[Jadenag]

Thanks for the help. Can someone also look into e and g since Iwe been told that those are incorrect. My answers for e and g are 27 and 3 respectively. Thanks.

 

[eumyang]

Looks like e) is right. You can also try using the difference of cubes formula
a³ - b³ = (a - b)(a² + ab + b²).
a, in this case, would be x^1/3.

 

[QuarkCharmer]

e looks good this time around.


Try graphing out g and check out what the function does as x approaches 1. In a situation like this you can break the limit into two cases and specify what the limit is as x approaches 1 from both the positive and negative side. Some professors simply call it undefined and leave it at that though.

 

[QuarkCharmer]

#f, L'Hopital's Rule gives these expressions,

First derivative of top and bottom:
$\frac{3x^2-7}{3x^2}$
Second derivatives...
$\frac{6x-0}{6x}$, which is clearly equal to 1 (that is, as x approaches 0).

After you take the first derivatives of the num/den there you no longer have an indeterminate form and cannot apply L'hopital's Rule. Unless there is some other case which allows the second derivative to be taken?

I don't believe that this limit is 1.

 

[Jadenag]

After you take the first derivatives of the num/den there you no longer have an indeterminate form and cannot apply L'hopital's Rule. Unless there is some other case which allows the second derivative to be taken?

I don't believe that this limit is 1.

If you get an indeterminate form when you take the ration of the derivatives of the two function you can consider the second and third derivatives as well.

 

[eumyang]

If you get an indeterminate form when you take the ration of the derivatives of the two function you can consider the second and third derivatives as well.

Instead of taking the 2nd derivative, take the 1st derivative:
$\lim_{x \rightarrow 0} \frac{3x^2 - 7}{3x^2}$
... and split into a difference of two fractions. Remember that the limit of a difference is the difference of the limits. See what happens from there.

 

[symbolipoint]

My efforts on #f are way off. Not yet sure what the best method is, but graphing the function shows moving toward negative infinity as x approaches 0. No limit.

 

[QuarkCharmer]

If you get an indeterminate form when you take the ration of the derivatives of the two function you can consider the second and third derivatives as well.

Yes but the problem in question did not result in an indeterminate form. The initial equation resulted in a type 0/0 ind. allowing L'hopital's rule to be applied. If you tried to take the limit again you would end up dividing by zero, but the numerator was a non-zero. I'm pretty sure that L'Hopital does not allow this by definition, as it is not an indeterminate form:

http://en.wikipedia.org/wiki/L'Hôpital's_rule#Other_indeterminate_formsTo solve limit f you might try taking the first derivative (L'hopital) as to end up with:

$$lim_{x \to 0} \frac{3x^{2}-7}{3x^{2}}$$

From there, I broke the fraction into parts:

$$1-\frac{7}{3x^{2}}$$

Now I simply factored out the 3 from the denominator and moved the limit on inside so that:

$$1- \frac{7}{3(lim_{x \to 0}x^2)}$$

Then just use the fact that $(lim_{x \to 0}x^{n}) = (lim_{x \to 0}x)^{n}$

Then you can just think about how when x goes to, but never "reaches" 0. the denominator get's really really small. Making the fraction tend to infinity. But it's one minus the fraction, so it shoots off to - infinity.

 

[uart]

My efforts on #f are way off. Not yet sure what the best method is, but graphing the function shows moving toward negative infinity as x approaches 0. No limit.

Use L'Hopital's once and the numerator is -7 while the denominator is zero. So yes, it doesn't reach a limit.

 

[Jadenag]

Yes but the problem in question did not result in an indeterminate form. The initial equation resulted in a type 0/0 ind. allowing L'hopital's rule to be applied. If you tried to take the limit again you would end up dividing by zero, but the numerator was a non-zero. I'm pretty sure that L'Hopital does not allow this by definition, as it is not an indeterminate form:

http://en.wikipedia.org/wiki/L'Hôpital's_rule#Other_indeterminate_forms


To solve limit f you might try taking the first derivative (L'hopital) as to end up with:

$$lim_{x \to 0} \frac{3x^{2}-7}{3x^{2}}$$

From there, I broke the fraction into parts:

$$1-\frac{7}{3x^{2}}$$

Now I simply factored out the 3 from the denominator and moved the limit on inside so that:

$$1- \frac{7}{3(lim_{x \to 0}x^2)}$$

Then just use the fact that $(lim_{x \to 0}x^{n}) = (lim_{x \to 0}x)^{n}$

Then you can just think about how when x goes to, but never "reaches" 0. the denominator get's really really small. Making the fraction tend to infinity. But it's one minus the fraction, so it shoots off to - infinity.

wow thankyou. And thankyou everyone else too!

 

[verty]

L'hopital's rule has three conditions, and if they don't apply, you can't use it. You must always check them.

(3x^2 - 7) / 3x^2 fails the first condition in the above question, the numerator and denominator do not have the same limit in that case.

 

[Jadenag]

L'hopital's rule has three conditions, and if they don't apply, you can't use it. You must always check them.

(3x^2 - 7) / 3x^2 fails the first condition in the above question, the numerator and denominator do not have the same limit in that case.

what are those three conditions?
As far as I am concerned all Iwe been doing is checking for an indeterminate form and if its tjere applying L hopitals rule.

Also. When you guys suggested breaking it into two parts and not going for the second derivative, I remember my teacher saying somewhere that if the ratio of the first derivatives of the two function is zero you consider the ratio of the second derivatives? Is that incorrect?

 

[dynamicsolo]

Try graphing out g and check out what the function does as x approaches 1. In a situation like this you can break the limit into two cases and specify what the limit is as x approaches 1 from both the positive and negative side. Some professors simply call it undefined and leave it at that though.

This is not described as two "cases" generally. What you describe is that the two "one-sided limits" do not agree, so the (two-sided) limit does not exist. It is not simply "undefined" because a standard "epsilon-delta" argument would fail for this function near x = 1.

Another way you can see that there's going to be trouble is that this function has an oblique asymptote of y = x + 1 and a vertical asymptote at x = 1 . For the graph to be contained by those lines, the curve for x > 1 must run off to "positive infinity" as x approaches 1 , while the branch for x < 1 must go to "negative infinity". So there is no meaning for a limit as x approaches 1.

 

[dynamicsolo]

what are those three conditions?
As far as I am concerned all Iwe been doing is checking for an indeterminate form and if its tjere applying L hopitals rule.

Also. When you guys suggested breaking it into two parts and not going for the second derivative, I remember my teacher saying somewhere that if the ratio of the first derivatives of the two function is zero you consider the ratio of the second derivatives? Is that incorrect?

If this is (f) you're talking about, you have to try something else because after taking first derivatives, the "limit" as x --> 0 is now "-7/0", so L'Hopital will avail you no further...

 

[QuarkCharmer]

This is not described as two "cases" generally. What you describe is that the two "one-sided limits" do not agree, so the (two-sided) limit does not exist. It is not simply "undefined" because a standard "epsilon-delta" argument would fail for this function near x = 1.

Another way you can see that there's going to be trouble is that this function has an oblique asymptote of y = x + 1 and a vertical asymptote at x = 1 . For the graph to be contained by those lines, the curve for x > 1 must run off to "positive infinity" as x approaches 1 , while the branch for x < 1 must go to "negative infinity". So there is no meaning for a limit as x approaches 1.

I haven't got into the epsilon-delta limit stuff. My professor used to have us break those limits into two cases, for when the variable is approaches fromt the positive and the negative. I'm sure you are probably right though.