Answer:
Step-by-step explanation:
Given that,
y=(1 + 1/n)ⁿ
As n goes larger 1/n approaches 0
y=1^∞
Then, this is the indefinite,
So let take 'In' of both sides
y=(1 + 1/n)ⁿ
In (y) =In ((1+1/n)ⁿ)
From law of logarithm
LogAⁿ=nLogA
Then, we have
In(y)= In ((1+1/n)ⁿ)
In(y)= n•In (1+1/n)
This can be rewritten to conform to L'Hospital Rule
In(y)= n / 1 / In(1+1/n)
As n approaches infinity
n also approaches infinity
And In(1+1/n) approaches 0, then, 1/In(1+1/n) approaches infinity
Then we have another indeterminate
Then, applying L'Hospital
Differentiating both the denominator and numerator
The differential of 1/In(1+1/n)
n^-2In(1+1/n) / (In(1+1/n))²
1 / n²In(1+1/n)
Then, apply L'Hospital
In(y) = 1/ n²In(1+1/n)
As n tends to infinity
In(y)= 0
Take exponential of both sides
y=exp(0)
y=1
As n goes larger the larger the irrational number but when n goes to infinity then the irrational number goes to 1
