Limits and Cont...

The value of $$\mathrm{f}({0})$$ so that the function $$\displaystyle \mathrm{f}({x})=\frac{\sqrt{1+x}-\sqrt[3]{1+x}}{x}$$ becomes continuous is equal to

If $$f(x)$$ is continuous and $$f\left(\dfrac {9}{2}\right)=\dfrac {2}{9}$$, then $$\displaystyle\lim _{ x\rightarrow 0 }f\left(\frac {1-\cos 3x}{x^2}\right)$$ is equal to :

If $$\displaystyle f(x) = \sqrt {\frac{{x - \sin x}}{{x + {{\cos }^2}x}}} $$ then $$\mathop {\lim }\limits_{x \to \infty } f(x)$$  is

$$\displaystyle\lim_{x \rightarrow \infty}(1^x + 2^x + 3^x+.........+n^x)^{1/x}$$ is

$$\displaystyle \lim_{x\to\infty}{\displaystyle \frac{(2x + 1)^{40}(4x - 1)^5}{(2x + 3)^{45}}}$$ is equal to

If $$f(x) =\left\{\begin{matrix}
\dfrac{8^{x}-4^{x}-2^{x}+1^{x}}{x^{2}},x>0 & \\
e^{x}\sin x+\pi x+\lambda \ln 4,x\leq 0 &
\end{matrix}\right.$$is continuous at $$x= 0$$, then $$\lambda$$ is a

The function $$f :( R-{0})$$ $$\rightarrow $$ R given by $$\displaystyle f(x)=\frac{1}{x}-\frac{2}{e^{2x}-1}$$ can be made continuous at $$x = 0$$ by defining $$f(0)$$ as

$$\displaystyle \lim_{n\to\infty}{\displaystyle \frac{n(2n + 1)^2}{(n + 2)(n^2 + 3n - 1)}}$$ is equal to

Let $$\mathrm{f}:\mathrm{R}\rightarrow \mathrm{R}$$ be defined by $$\mathrm{f}(\mathrm{x})=\left\{\begin{array}{l}
\mathrm{k}-2\mathrm{x}, \mathrm{i}\mathrm{f}   \mathrm{x}\leq-1\\
2\mathrm{x}+3, \mathrm{i}\mathrm{f}   \mathrm{x}>-1
\end{array}\right.$$ be continous. then find possible value of $$\mathrm{k}$$ is

If $$f(x) = \displaystyle \left\{\begin{matrix}x - 1, & x \geq 1 \\ 2x^2 - 2, & x < 1\end{matrix}\right. , g(x) = \left\{\begin{matrix}x + 1, & x > 0 \\ -x^2 + 1, & x \leq 0\end{matrix}\right.$$, and $$h(x) = |x|$$, then $$\displaystyle \lim_{x \rightarrow 0} f(g (h (x)))$$ is