જો $\mathop {\lim }\limits_{x \to a} \frac{{{x^9} + {a^9}}}{{x + a}} = 9$ હોય,તો $a = $

કિંમત શોધો: $\mathop {\lim }\limits_{x \to 0} \frac{\sqrt{1+x}-1}{x}$

$\mathop {\lim }\limits_{\theta \to \pi /6} \frac{{\cot^2 \theta - 3}}{{\csc \theta - 2}} = $

જો $a > 0$ હોય,$[\cdot]$ એ મહત્તમ પૂર્ણાંક વિધેય દર્શાવે છે,$\lim _{x \rightarrow a^{-}}\left(\frac{|x|^3}{a}-\left[\frac{x}{a}\right]^3\right)=k$,અને $\lim _{x \rightarrow a^{+}}\left(\frac{|x|^3}{a}-\left[\frac{x}{a}\right]^3\right)=l$,તો:

$\lim _{x \rightarrow 0} \frac{\sqrt{1-\cos x^2}}{1-\cos x} = $

$\mathop {{\rm{lim}}}\limits_{x \to 0} \frac{{\left( {1 - \cos 2x} \right)\left( {3 + \cos x} \right)}}{{x\tan 4x}} = $