Limits and Cont...

Evaluate: $$\displaystyle \underset{x\rightarrow 0}{\lim}\ \ \frac{\sin3x^{2}}{\cos(2x^{2}-x)}$$

lf $$f(x)=\displaystyle \begin{cases}\dfrac{a^{2[x]+\{x\}}-1}{2[x]+\{x\}};x\neq 0 \\ \log a;x=0 \end{cases}$$ where $$[.\ ]$$ and $$\{.\ \}$$ denote integral and fractional part respectively, then

 lf $$\displaystyle { f }({ x })=\sqrt { \frac { { x }-\sin ^{ 2 }{ x }  }{ { x }+\cos { x }  }  } $$,then $$\displaystyle \lim _{ x\rightarrow \infty  } f(x)$$=

The value of $$\displaystyle \lim _{ x\rightarrow 0 }{ \frac { \sin { \left( \pi \cos ^{ 2 }{ x }  \right)  }  }{ { x }^{ 2 } }  } $$ is

Let $$f(x)=\cos2x.\cot\left (\displaystyle  \frac{\pi }{4}-x \right )$$ If $$f$$ is continuous at $$x=\displaystyle \frac{\pi}{4}$$ then the value of $$f(\displaystyle \frac{\pi}{4})$$ is equal to

$$\displaystyle \lim_{x\rightarrow \infty }(\sin\sqrt{x+1}-\sin\sqrt{x})=$$

$$\displaystyle \lim_{x\rightarrow \infty }x\displaystyle \cos\left(\frac{\pi}{8x}\right)\sin\left(\frac{\pi}{8x}\right)=$$

 $$\displaystyle \lim_{x\rightarrow\infty}\frac{\sin^{4}x-\sin^{2}x+1}{\cos^{4}x-\cos^{2}x+1}$$ is equal to

$$f(x)=\left\{\begin{matrix}[x]+[-x], & \\  \lambda ,& \end{matrix}\right.\begin{matrix}x\neq 2 & \\  x=2& \end{matrix},$$ then f(x) is continuous at $$x=2$$ provided $$\lambda $$ is:

$$\displaystyle \lim_{x\rightarrow \infty }2^{-x}\sin(2^{x})$$