If `lim_(x->oo)(x^2+3x+5)/(4x+1+x^k)` exists then `k=2` (b) `k<2` (c) `k >2` (d) `k >=2`

If lim_(x->oo)(x^2+3x+5)/(4x+1+x^k) exists then K (a) k =2 (b) k 2 (d) k >=2

If lim_(x rarr oo)(x^(2)+3x+5)/(4x+1+x^(k)) exists then k=2(b)k 2(d)k>=2

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

If (x-4)/(x^(2)-5x-2k)=2/(x-2)-1/(x+k), " then "k=

If (x-4)/(x^(2)-5x-2k)=(2)/(x-2) - (1)/(x+k) , then find k ?

If (x-4)/(x^(2)-5x+2k)=(2)/(x-2) - (1)/(x-k) , then find k ?

If lim_(x to 2)(tan(x-2){x^(2)+(k-2)x-2k})/(x^(2)-4x+4)=5 then k is equal to

If lim_(x rarr1)cos ec^(-1)((k^(2))/(ln x)-(k^(2))/(x-1)) exists, then k belongs to the interval exists,

Find k so that lim_(x rarr 2) f(x) exists, where f(x) = {(2x+3 if xle2),(x + k if x gt2):}

If lim_(x to 2) (x - 2)/(""^(3)sqrt(x) - ""^(3)sqrt(2)) = lim_(x to k) (x^(2) - k^(2))/(x - k) find the value of K