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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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Answer» `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 `=lim_(x-0) (2e^(2x)xx2-0)/2(xe^(2x)xx2+e^2x)+e^2xxx2-0` `=lim_(x-0)(4e^2x)/4xe^2x+4e^2x` `=4/4` option(4)`=1` |
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