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