- #1

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## 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 doesnt seem like a very good answer, so i dont 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'hopitals and got:

lim x->0+ [(2(sec2x)^2)/tan2x] / -1/x^2

I kept getting indeterminate forms and reapplying l'hopitals 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 realise 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 isnt 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 im 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 :)