l'hospitals rule, a problem?

Question

what's the limit as x approaches infinity when

x/(ln(1+2e^x))

Answer 1

L'hopital's rule also applies to infty/infty forms and so can be used in this problem. As others have pointed out, the limit is 1.

However, it is usually much better to do limits without L'Hopitals's rule, either by looking at dominant terms of the functions involved or by doing a power series expansion. In this case, ln(1+2e^x) is about ln(2e^2), which is about ln(2)+x. Now, x/[ln(2)+x] clearly goes to 1. In practice, this type of reasoning is more versatile than L'Hopital's rule.

Answer 2

L'hospital's rule is NOT needed, and in fact is QUITE INAPPROPRIATE. ###

The limit is 1.

This is the answer Christina C gave, but her reasoning is false and simply doesn't apply to the function that you posed.

All you first need to do is to find the limiting functional form of the denominator when x becomes very large. For large x,

ln(1+2e^x) ---> ln(2e^x) = ln 2 + x = x (1 + [ln 2] / x)

[I'm ignoring an even smaller and less significant correction term due to the original ' 1 ' inside the logarithm.]

So for LARGE x, x / (ln(1+2e^x)) ---> x / {x (1 + [ln 2] / x)}

---> 1 / (1 + [ln 2] / x) ---> 1.

Live long and prosper.

### L'hospital's rule applies ONLY to functions of the form that appear to be indeterminate for some value of the independent variable x because both numerator and denominator are simultaneously ZERO for that value of x. That is, before analysing them further, it appears that they take on the apparent "value" 0/0 at some x = x_0.

One could only FORCE this function into the form where applying L'hospital's would be appropriate in the most ARTIFICIAL way, as follows:

Let y = 1 / x. Consider the function

F(y) = A(y) / y, where the numerator A(y) itself is given by:

A(y) = 1 / [ln (1 + 2 e^(1/y)] !!

NOW you have both the numerator, A(y), and the denominator, y, tending to zero as y ---> 0. Thus the conditions for L'hospital's rule are now met.

But as I said, that's an EXTREMELY ARTIFICIAL FORCING into a form neither desirable NOR necessary. You can just LOOK at the expression as a function of x and SEE what the limit will be for large x, without doing anything remotely fancy or ingenious.

Answer 3

By itself you see the limit seems to go to ∞/∞, so you can use l'Hôpital's rule here (somebody claimed it's only for 0/0, which isn't true. See the reference link below). So l'Hôpital's rule says you can take the derivative of the top and bottom. You get "1" on top and (1 / (1+2e^x))*2e^x for the bottom. So the new expression is (1+2e^x) / 2e^x. This gives you ∞/∞, so you can use l'Hôpital's rule again and get 2e^x / 2e^x, which is just 1. So the limit is 1.

Alternitively, on the next to last step you could rewrite things as (1 / 2e^x)+(2e^x / 2e^x) = (1 / 2e^x) + 1 , and the limit of this goes to 0+1 = 1

Answer 4

Hey, Spock, I think you're being a little tough on Christina. L'Hospital's Rule can be used if both numerator and denomator approach infinity as well (not just zero). IMO the application of L'Hospital's Rule here is okay.

Answer 5

limit=1

if you keep applying L'Hospitals, you will eventually get (2e^x)/(2e^x)=1 and the limit as x goes to infinity of 1 is 1.

Answer 6

differentiate numer & denom ...this u can do until any of it is zero...when it is becoming zero u have to stop it... here the answer is 2
(
apply l'hospital rule u get 1/e^x/(1+2e^x)
= (1+2e^x)/e^x
= 1/e^x + 2
=> apllying lt we get 2
)

Answer 7

lim(x/(ln(1+2e^x))) =
x→∞
lim(1/(2e^x)/(1 + 2e^x)) =
x→∞
lim((1 + 2e^x)/(2e^x)) = 1
x→∞