Limits and Cont...

Let $$f : R \to R$$ be a differentiable function satisfying $$f'(3) + f'(2) = 0$$.
Then $$\underset{x \to 0}{\lim} \left(\dfrac{1+f(3+x)-f(3)}{1+f(2-x) - f(2)}\right)^{\frac{1}{x}}$$ is equal to 

Let $$f$$ be a differentiable function such that $$f'(x) = 7- \dfrac{3}{4}\dfrac{f(x)}{x}, (x > 0)$$ and $$f(1) \neq 4$$.
Then $$\underset{x\to 0^+}{\lim} xf \left(\dfrac{1}{x}\right) $$:

If $$f(x) = \begin{vmatrix} \cos x& x & 1\\ 2\sin x & x^{2} & 2x\ \\ \tan x & x & 1\end{vmatrix}$$, then $$\displaystyle \lim_{x\rightarrow 0} \dfrac {f'(x)}{x}$$.

$$\lim _{ { x\rightarrow \pi /4 } } \dfrac { \cot ^{ { 3 } } x-\tan  x }{ \cos  (x+\pi /4) } $$  is

If the function $$f(x)= \left\{\begin{matrix}\dfrac {\sqrt {2+\cos x}-1}{(\pi-x)^2} & x\neq \pi \\ k & x=\pi \end{matrix}\right.$$ is continuous at $$x=\pi$$, then $$k$$ equals :

$$\underset{x\to 0}{\lim} \left(\dfrac{3x^2+2}{7x^2+2}\right)^{1/x^2}$$ is equal to:

If $$\lim _{ x\rightarrow \infty  }{ x\sin { \left( \cfrac { 1 }{ x }  \right)  }  } =A$$ and $$\lim _{ x\rightarrow 0 }{ x\sin { \left( \cfrac { 1 }{ x }  \right)  }  } =B$$, then which one of the following is correct?

$$\displaystyle \lim_{x\rightarrow 0}x^{2}\displaystyle \sin\frac{\pi}{x}=$$

If $$x$$ is very large, then $$\dfrac {2x}{1+x}$$ is

$$\displaystyle \lim_{x\rightarrow \infty} \sin x$$ equals