If `k x^2+2`, for x at `x^4 +1`, for `x >0` equation : `f(x)= lim (k(x^2-2) =k,lim (x^2-2)`

lim_(x->1 (k^2/logx -k^2/(x-1))

In each of the following, determine the value of k for which the given value is a solution of the equation: (i) k x^2+2x-3=0,\ \ x=2 (ii) 3x^2+2k x-3=0,\ \ x=-1/2 (iii) x^2+2a x-k=0,\ \ x=-a

lim_ (x rarr-oo) x (sqrt (x ^ (2) + k) -x)

Statement 1: lim_ (x rarr oo) ((1 ^ (2)) / (x ^ (3)) + (2 ^ (2)) / (x ^ (3)) + (3 ^ (2)) / (x ^ (3)) + ...... + (x ^ (2)) / (x ^ (3))) = lim_ (x rarr oo) (1 ^ (2)) / (x ^ ( 3)) + lim_ (x rarr oo) (2 ^ (2)) / (x ^ (3)) + ...... + lim_ (x rarr a) (x ^ (2)) / (x ^ (3)) lim_ (x rarr a) (f_ (1) (x) + f_ (2) (x) + ... + f_ (n) (x)) = lim_ (x rarr a) f_ (1) (x) + lim_ (x rarr a) f (x) + ...... + lim_ (x rarr a) f_ (n) (x)

Find the value of k, if lim_(x rarr 1) (x^(4)-1)/(x-1) = lim_(x rarr k)(x^(3)-k^(3))/(x^(2)-k^(2))

Find k so that : lim_(x rarr 2)f(x) exists, where : f(x)={(2x+3,"," if xle2), (x+2k,"," if x>2.):}

Find k so that : lim_(x rarr 2)f(x) exists, where : f(x)={(2x+3,"," if xle2), (x+k,"," if x>2.):}