Use the substitution
Standard limit:
Here's the limit I'm working with:
Now, I understand that the correct answer is . While I don't fully understand how you get there, I understand that you can manipulate this until you come up with .
However, here's how I'm looking at it...
As x approaches infinity, the expression within the parentheses approaches 1. Which to me, means that the limit of this whole thing is , which is one.
So, my question is - how is the way I'm conceptually looking at this incorrect?
You have an indeterminate form!
Just like
So we need to resolve this limit in a different way
Since all of the functions are continuous and the natural log and exponential are monotone increasing we can use this identity
Now this is of the form zero times and infinity and can be put in the form 0 over 0
Now by L'hopitals rule we get
The base is approaching 1 but is always a little bigger than 1 and the exponent is going to infinity. Loosely speaking it is like a race does the base go to 1 faster then the exponent goes to infinity.
Indeterminate form - Wikipedia, the free encyclopedia
Well, where exactly do you want to start? Do you accept that the derivative of e^x is e^x? Then you accept that lin_(x->0) (e^x- 1)/x= 1. That is, for x close to 0, (e^x- 1)/x is approximately 1 so e^x- 1 is approximately x and then e^x is approximately x+ 1. From that, x is approximately ln(x+ 1). From that ln(x+1)/x= ln(x+ 1)^{1/x} is approximately 1. Then (x+ 1)^{1/x} is approximately e.
Now, let n= 1/x so that x close to 0 becomes n very large: for very large n, (1+ (1/n))^n is approximately e and as n goes to infinity, lim_{n->infinity} (1+ (1/n))^n= e.