How do you find the limit of #(x/(x+1))^x# as x approaches infinity?

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Answer 1 · 2016-09-24T09:35:18

#"The Reqd. Limit="1/e.#

We will use the Standard Limit # : lim_(yrarr0) (1+y)^(1/y)=e#.

Let, #x=1/y", so that, as "xrarroo, yrarr0#.

Now, the Reqd. Limit #=lim_(xrarroo)(x/(x+1))^x#

#=lim_(yrarr0){(1/y)/(1/y+1)}^(1/y)#

#=lim_(yrarr0){1/(1+y)}^(1/y)#

#=lim_(yrarr0)(1+y)^(-1/y)#

#=lim_(yrarr0){(1+y)^(1/y)}^-1#

Since, the Power Function # :trarrt^-1# is continuous, we have,

#"The Reqd. Limit="{lim_(yrarr0)(1+y)^(1/y)}^-1=e^-1=1/e.#

Enjoy Maths.!