`lim_(x->2) f(x)`, f(x) = `{(x - [x] , x < 2) , (4 , x=2) , (3x-5 , x > 2 ):}`

lim_(x rarr2)(f(x)-f(2))/(x-2)=

lim_(x rarr1)f(x), where f(x)={x^(2)-1,x 1

Find lim_(x to 0)f(x) and lim_(x to 1)f(x) , where f(x)={{:(2x+3",", x le 0), (3(x+1)",", x gt 0):}

If f (x) = x^(4) + 2x^(3) , them lim_(x to 2) (f(x) - f(2))/(x - 2) is equal to

If lim_( xto 2 ) (f(x) -f(2))/( x-2) exist ,then

Calculate lim_(x to 0) f(x) , where f(x) = (1)/(x^(2)) for x gt 0

If f(x)=|x-2| ,then (a) lim_(x rarr2+)f(x)!=0 (b) lim_(x rarr2-)f(x)!=0 (c) lim_(x rarr2+)f(x)!=lim_(x rarr2-)f(x) (d) f(x) is continuous at x=2

If (lim)_(x->2)(f(x-9))/(x-2)=3 then (lim)_(x->2)f(x), is