Let `f: RvecR` be such that `f(a)=1,f^(prime)(a)=2.` Then `(lim)_(xvec0)((f^2(a+x))/(f(a)))^(1//x)` is a. `e^2` b. `e^4` c. `e^(-4)` d. `1//e`

Let f:Rvec R be such that f(a)=1,f'(a)=2 Then (lim_(x))_(x0)((f^(2)(a+x))/(f(a)))^(1/x) is a e^(2) b.e^(4) c.e^(-4)d.1/e

Let f: R->R be such that f(1)=3 and f^(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

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

Let f:R rarr R be such that f(1)=3andf'(1)=6. Then lim_(x rarr0)((f(1+x))/(f(1)))^(1/x)=(a)1( b) e^((1)/(2))(c)e^(2) (d) e^(3)

Let f : R to R ,f (a) =1, f '(a) = 2 then prove that Lt _(x to 0) ((f ^(2) (a + x ) )/( f (a)))^(1/x) = e ^(4)

Let R rarr R be such that f(1)=3 and *f'(1)=6. then lim_(x rarr0)((f(1+x))/(f(1))) equal : (1)1(2)e^((1)/(2))(3)e^(2) (4) e^(3)

Let f(x) be the fourth degree polynomial such that f^(prime)(0)-6,f(0)=2a n d(lim)_(xvec1)(f(x))/((x-1)^2)=1 The value of f(2) is 3 b. 1 c. 0 d. 2

Let f(x) be a function such that ("lim")_(xvec0)(f(x))/x=1 and ("lim")_(xvec0)(x(1+acosx)-bsinx)/((f(x))^3)=1 , then b-3a is equal to

Let f(x) be a twice-differentiable function and f"(0)=2. The evaluate: ("lim")_(xvec0)(2f(x)-3f(2x)+f(4x))/(x^2)

If f: RvecR be a differen tiable function atg x=0 satisfying f(0)=0a n df^(prime)(0)=1 , then the value of (lim)_(xvec0)1/xsum_(n=1)^oo(-1)^nf(x/n)= e b. -log2 c. 0 d. 1