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If the function f: Rsetminus {0}vec give...

If the function `f: Rsetminus` `{0}vec` given by `f(x)=1/x-2/(e^(2x)-1)` is continuous at `x=0,` then find the value of `f(0)`

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`F(x)=1/x-2/(e^2x-1)`
`lim_(x-6^+)F(x)=lim_(x+6^-)F(x)=F(0)`
`lim_(x-0)F(x)=lim_(x+0)(1/11-2/(e^(2x)-1))`
`=lim_(x-0) (e^(2x)-1-2x)/(x(e^2x-1))`
`=(0/0)L`hospital
`=lim_(x-0) (e^(2x)xx2-0-2)/(x(e^(2x)xx0-0)+e^(2x))`
`=lim_(x-0)(2e^(2x)-2)/(2xe^2x+e^2x-1)`
`(0/0)L`hospital
...
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