For `x in R`, `lim_(xrarroo)((x-3)/(x+2))^x` is equal to (a) `e` (b) `e^(-1)` (c) `e^(-5)` (d) `e^5`

("lim")_(xrarroo)(1/e-x/(1+x))^x is equal to (a) e/(1-e) (b) 0 (c) e/(e^(1-e)) (d) does not exist

lim_(nrarroo) (1-x+x.root n e)^(n) is equal to

lim_(x->oo) (e^(1/x^2)-1)/(2tan^-1(x^2)-pi) is equal to (a) 1 (b) -1 (c) 1/2 (d) -1/2

' lim_ (x to 0) (a^(x)-b^(x))/(e^(x)-1) is equal to

If lim_(xrarroo) ((x+c)/(x-c))^(x)=4 then the value of e^(c) is

lim_(xto0)(e^(x)-e^(-x))/sinx

lim_(x->oo)[sqrt(x+sqrt(x+sqrt(x)))-sqrt(x)] is equal to (a)0 (b) 1/2 (c) log 2 (d) e^4

The value of lim_(x rarr 1)(2-x)^(tan((pix)/2)) is (a) e^(-2/pi) (b) e^(1/pi) (c) e^(2/pi) (d) e^(-1/pi)

lim_(xto1)(e^(x)-e)/(x-1)= ……………

Let f: R->R be such that f(1)=3a n df^(prime)(1)=6. Then lim_(x->0)((f(1+x))/(f(1)))^(1//x)= (a) 1 (b) e^(1/2) (c) e^2 (d) e^3