Prove that `ln (1+1/x) gt (1)/(1+x), x gt 0`. Hence, show that the function `f(x)=(1+1/x)^(x)` strictly increases in `(0, oo)`.

Explanation

`f(x) -(1+(1)/(x))^(x) " for Domain of "f(x) 1+(1)/(x) lgt0`
`rArr (x+1)/(x) gt0 rArr (-oo , -1) uu (0,oo)`
`"Consider "f(x) =(1+(1)/(x))^(2) [en (1+(1)/(x))+(x)/(1+(1)/(x)).(-1)/(x^(2))]`
`rArr f(x) =(1+(1)/(x))^(x) [en (1+(1)/(x))-(1)/(x+1)]`
Now `(1+(1)/(x))^(x)` is always positive , hence the sigh of f(x) depends on sigh of `en (1+(1)/(x)) -(1)/(1+x)` i.e., we have to compare en `(1+(1)/(x))` and So lets assume g (x) `=en (1+(1)/(x))- (1)/(x+1)`
`g(x) =(1)/(1+(1)/(x)).(-1)/(x^(2)) + (1)/((x+1)^(2)) " "rArr " " g(x) =(-1)/(x(x+1)^(2))`
`(i) " ""for " x in (0,oo), g(x) lt 0 " " rArr " " g(x) " is " M.D " for " x in (0,oo)`
`g(x) lt underset(x to oo)("lim") g (x)`
`g(x) gt 0 " ""and since " g (x) gt0 " " rArr " " f(x) gt 0`
`(ii) " ""for " x in (-oo ,-1) g(x) gt0 " " rArr " " g(x) " is " M.I. "for " x in (-oo , -1)`
`rArr " " g(x) gt underset( x tooo)("lim") g (x) " " rArr " " g(x) gt0 " " rArr " " f(x) gt 0`
Hence from (i) and (ii) we get f(x) `gt` 0 for all `x in (-oo ,-1) uu (0,oo)`
`rArr " "f(x)` is M.I. in its Domain
For drawing the graph of f(x) its important to find the value of f(x) at boundary points i.e., `+- oo ,0 ,-1`
`underset( x tooo)("lim") (1+(1)/(x))^(x) =e`
`underset(x to0)("lim") (1+(1)/(x))^(x) =1 " " and " "underset(x to-1)("lim") (1+(1)/(x))^(x)=oo`

so the graph of f(x) is
Range is ` y in (1,oo) -{e}`