The value that (1 plus StartFraction 1 Over n EndFraction )Superscript n1 1 nn approaches as n gets larger and larger is the irrational numberTrue/False

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

answer: idk

step-by-step explanation: