Indeterminate Limits (L'Hopitals Rule)

[Jet1045]

Homework Statement

Hey Everyone , i have 4 L'Hopital's rule problems that i am totally stuck on, and hints would be greatly appreciated. Ill just put all 4 here, and any help would be great.

1. lim x->a+ (cos(x)ln(x-a)) / ln(e^x - e^a)

Okay, so with this one i determined that it has the form ∞/∞. So i took the deriv of top and bottom.

and i got lim x->a+ [-sin(x)ln(x-a) + cos(x)/(x-a)] / [(e^x) / ( (e^x)-(e^a)].
When i substitute 0 i got ∞/0 . Which doesn't seem like a very good answer, so i don't know where i went wrong with my derivative.

2. lim x->0+ (tan2x)^x

This one would of course be in the form 0^0. So i know i need to so e^ lim x->0+ xln(tan2x). Which has the form 0(0) so i need to convert the indeterminate product into a quotient.
So i make lim x->0+ ln(tan2x)) / 1/x. This is form 0/0 so i applied l'hospital's and got:
lim x->0+ [(2(sec2x)^2)/tan2x] / -1/x^2

I kept getting indeterminate forms and reapplying l'hospital's but it seems to be getting no where, so i obviously messed up in the beginning and i can't see where.

3. lim x->∞ (e^x + x)^(1/x)

I realize that this problem is in the form ∞^0.
So i rewrote it as x->∞ ln(e^x + x)/x which has the type ∞/∞ so applied LH.
And got [e^x / (e^x + x)] / 1. When i substitue in i get ∞/1 Which isn't an indeterminate form, so is the answer just ∞. I know i would then have a final answer of e^∞. Which would be ∞.
This question is near the end of the textbook section, so i assumed it would be more difficult, but some confirmation here would be nice.

4. lim x->∞ [(2x-3)/(2x+5)]^(2x+1)
I treated this as an indeterminate power and got a whole lot of messy expansions, and such, so i think I am wayyy on the wrong track. But since the original limit would be ∞^∞ Is this still an indeterminate form? In class we only went over three indeterminate powers: 0^0 , ∞^0, and 1^∞. So I'm not even sure if this would be in an indeterminate form. Any help with this one would be great cause I'm totally lost.

If anyone, can help me on any of these i'd appreciate it. Thanks :)

 

[SammyS]

Homework Statement

Hey Everyone , i have 4 L'Hopital's rule problems that i am totally stuck on, and hints would be greatly appreciated. Ill just put all 4 here, and any help would be great.

1. lim x->a+ (cos(x)ln(x-a)) / ln(e^x - e^a)

Okay, so with this one i determined that it has the form ∞/∞. So i took the derivative of top and bottom.

and i got lim x->a+ [-sin(x)ln(x-a) + cos(x)/(x-a)] / [(e^x) / ( (e^x)-(e^a)].
When i substitute 0 i got ∞/0 . Which doesn't seem like a very good answer, so i don't know where i went wrong with my derivative.

...

If anyone, can help me on any of these I'd appreciate it. Thanks :)

****************
Added in Edit:

I don't much like how the following turned out. It certainly looks like a fairly complicated problem.​

*****************

I'll start by responding to the first one.

Simplify your result.

\displaystyle \frac{-\sin(x)\ln(x-a) + \displaystyle \frac{\cos(x)}{x-a}}{\displaystyle \frac{e^x}{e^x-e^a}}

\displaystyle =\frac{\left(-\sin(x)\ln(x-a) + \displaystyle \frac{\cos(x)}{x-a}\right)\left(e^x-e^a\right)}{e^x}\quad\quad\quad\quad\quad\quad \text{#1}


\displaystyle =\frac{\left\{-(x-a)\sin(x)\ln(x-a) +\cos(x)\right\}\left(e^x-e^a\right)}{(x-a)e^x}\quad\quad\quad \text{#2}
​

The limit of expression #1 is of the form \displaystyle \frac{\pm(\infty\pm\infty)\cdot 0}{e^a}\,. We don't know what the value of the parameter, a, is, so we don't know the signs of sin(a) and cos(a).

The limit of expression #2 is of the form \displaystyle \frac{(\pm0\cdot\infty+\cos(a))(0)}{0}\,.

The above forms assume that sin(a) ≠ 0, cos(a) ≠ 0

In any case, you need to put your expression in a form suitable to use L'Hôpital's rule again.

 

[SammyS]

...

2. lim x->0+ (tan2x)^x

This one would of course be in the form 0^0. So i know i need to so e^ lim x->0+ xln(tan2x). Which has the form 0(0) so i need to convert the indeterminate product into a quotient.
So i make lim x->0+ ln(tan2x)) / 1/x. This is form 0/0 so i applied l'hospital's and got:
lim x->0+ [(2(sec2x)^2)/tan2x] / -1/x^2

I kept getting indeterminate forms and reapplying l'hospital's but it seems to be getting no where, so i obviously messed up in the beginning and i can't see where.
...

Actually lim _x→0+ (x)ln(tan(2x)) is of the form 0∙(-∞) .

Take your result, \displaystyle \frac{2\sec^2(2x)}{-\tan(2x)(1/x^2)}\,, change sec(2x) to 1/cos(2x) and tan(2x) to sin(2x)/cos(2x) and multiply by x²/x² & simplify.

Apply L'Hôpital's rule one more time.

 

[SammyS]

3. lim x->∞ (e^x + x)^(1/x)

I realize that this problem is in the form ∞^0.
So i rewrote it as x->∞ ln(e^x + x)/x which has the type ∞/∞ so applied LH.
And got [e^x / (e^x + x)] / 1. When i substitute in i get ∞/1 Which isn't an indeterminate form, so is the answer just ∞. I know i would then have a final answer of e^∞. Which would be ∞.
This question is near the end of the textbook section, so i assumed it would be more difficult, but some confirmation here would be nice.

I haven't looked at your solution in detail, but I get a different answer.

Try letting u = 1/x, and take the limit as u→0⁺

Added in Edit:

[e^x / (e^x + x)] / 1 is not of the form ∞/1.

[e^x / (e^x + x)] / 1 = [e^x / (e^x + x)] which is of the form ∞/∞.

Use L'Hôpital's rule again.

 

[SammyS]

...

4. lim x->∞ [(2x-3)/(2x+5)]^(2x+1)
I treated this as an indeterminate power and got a whole lot of messy expansions, and such, so i think I am wayyy on the wrong track. But since the original limit would be ∞^∞ Is this still an indeterminate form? In class we only went over three indeterminate powers: 0^0 , ∞^0, and 1^∞. So I'm not even sure if this would be in an indeterminate form. Any help with this one would be great cause I'm totally lost.

This is of the form (1)^∞ . That might remind you of lim (1+(1/n))ⁿ as n→∞ , which is one of the definitions of e, (Euler's constant).

You should be able to do it with e raised to the natural log of the limit. Alternatively, it can be done by using the definition of e .

 

[SammyS]

1. lim x->a+ (cos(x)ln(x-a)) / ln(e^x - e^a)

For this, I suggest letting u = x-a then, x = u + a .

Take the limit as u → 0⁺.

Use the angle addition identity for cos(u+a).

The result may be quite different if a is a multiple of π/2 . --- or perhaps an odd multiple of π/2 .