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$\underset{x\to 0}{\mathop{\lim }}\,\left[ \frac{\sin (x+a)+\sin (a-x)-2\sin a}{x\sin x} \right]=$
$\underset{x\to 0}{\mathop{\lim }}\,{{\left( \frac{1+5{{x}^{2}}}{1+3{{x}^{2}}} \right)}^{1/{{x}^{2}}}}=$
$\underset{x\to \infty }{\mathop{\lim }}\,\frac{(2x-3)(3x-4)}{(4x-5)(5x-6)}=$
If $f(x)=\frac{\sin ({{e}^{x-2}}-1)}{\log (x-1)},$ then $\underset{x\to 2}{\mathop{\lim }}\,f(x)$ is given by
$\underset{x\to \infty }{\mathop{\lim }}\,(\sqrt{{{x}^{2}}+8x+3}-\sqrt{{{x}^{2}}+4x+3})=$
If $\underset{x\to 5}{\mathop{\lim }}\,\frac{{{x}^{k}}-{{5}^{k}}}{x-5}=500$ , then the positve integral value of k is
$\underset{x\to 0}{\mathop{\lim }}\,\frac{\sqrt{1-{{x}^{2}}}-\sqrt{1+{{x}^{2}}}}{{{x}^{2}}}$ is equal to
If $f(x) = \begin{cases} x, \text{if x is rational }\\ - x,\text{if x is irrational} \end{cases}$, then $\underset{x\to 0}{\mathop{\lim }}\,f(x)$ is
$\underset{x\to 0}{\mathop{\lim }}\,\frac{x{{e}^{x}}-\log (1+x)}{{{x}^{2}}}$ equals
The value of $\underset{x\to -\infty }{\mathop{\lim }}\,\frac{\sqrt{4{{x}^{2}}+5x+8}}{4x+5}$ is

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