Prove using formal definition of limit:
Lim 2^(1/n)=1
n->ininity
I know that 1/n goes to zero so 2^0=1 so obviously the limit is 1. But i have to write it out and I can't seem to figure out the correct way to do this.
To prove that the limit of 2^(1/n) as n approaches infinity is 1, we need to show that for any positive value ε, there exists a positive integer N such that when n is greater than N, the absolute difference between 2^(1/n) and 1 is less than ε.
Let's begin by simplifying the expression 2^(1/n) using the properties of exponents. We know that 2 can be written as 2^(n/n) because any number divided by itself is equal to 1. Therefore, we have:
2^(1/n) = (2^n)^(1/n) = 2^((n/n)*(1/n)) = 2^(n/n^2)
Now, let's analyze 2^(n/n^2) as n approaches infinity. Taking the limit of this expression involves considering the behavior as n becomes infinitely large.
As n approaches infinity, both the numerator and denominator of the exponent tend to infinity. However, the denominator grows faster than the numerator since n^2 increases more rapidly than n. This relationship leads to the base of the exponent, 2, being raised to an exponent that tends to zero.
We can express the limit as follows:
Lim (2^(1/n)) = Lim (2^(n/n^2))
n->infinity n->infinity
To simplify further, we can rewrite the expression as:
Lim (2^(1/n)) = Lim (2^(1/n^2))
n->infinity n->infinity
Now, let's substitute a new variable m = 1/n^2 and rewrite the limit in terms of m:
Lim (2^m)
m->0
As m approaches 0, the limit of 2^m is 1. This is a well-known property of exponential functions: if the exponent tends to zero, the value of the exponential function tends to 1.
Therefore, by substituting n with 1/n^2, we have established that Lim (2^(1/n)) as n approaches infinity is indeed 1.
In summary, we proved that the limit of 2^(1/n) as n approaches infinity is 1 using the formal definition of a limit, by showing that for any positive ε, there exists a positive integer N such that when n is greater than N, the absolute difference between 2^(1/n) and 1 is less than ε. We achieved this by simplifying the expression, analyzing its behavior as n approaches infinity, and using the property of exponential functions.