limit of (1+2/n)^n || step by step Explanation: convergence and divergence

Today we will learn how we can find the limit of (1+2/n)^{^n} in an easy way and in steps. In calculus, the limit of a function is an important topic of mathematics. From that, we have taken to find the limit of (1+2/n)^{^n} step by step.

How to find the limit of (1+2/n)^{^n} as n goes to infinity?

To find the limit of (1+2/n)^n as n goes to infinity you need to follow the following steps.

Step 1:

a_n=(1+2/n)^ⁿ

Step 2:

Take limit n→∞   on both sides

lim  (n→∞)  a_n                 = lim  (n→∞ )     (1+2/n)^{^n}

                                

We know that,

 lim  (n→∞)      [(1+b/n)ⁿ= e^{^b}]

                                

Hence,            lim  (n→∞)      (1+2/n)^{^n} will be

                           

lim  n→∞   e^{^2} ANS

Whether (1+2/n)^{^n} is Convergence and Divergence?

To find whether (1+2/n)^{^n} convergence and divergence. The simple is answer is that (1+2/n)^{^n} is convergent because the limit of (1+2/n)^{^n}≠∞

familiar questions with the limit  (1+2/n)^{^n} as n go to infinity.

We will study some familiar questions with the limit (1+2/n)^n. here is a question (1+1/n)^{^6n}. We will show you how to solve (1+1/n)^{^6n} by following the method of (1+2/n)^{^n}.

Take limit n→∞       on both sides

lim  (n→∞) a_n     = lim  (n→∞) (1+1/n)^{^6n}

                          

We know that,  

lim  (n→∞)     [(1+y/n)ⁿ= e^{^y}]

                               

 write, (1+1/n)^{^6n}    as [(1+1/n)^{^n}]^{^6}

Hence,  lim  (n→∞)   [(1+1/n)^{^n}]^{^6} will be

                            

lim  n→∞   e^{^6} ANS

See Also: Prove that lim  n→∞ (1+1/n)^mn=e^m