લક્ષ શોધો: $\mathop {\lim }\limits_{x \to 3} [x(x+1)]$

  • A
    $10$
  • B
    $11$
  • C
    $12$
  • D
    $13$

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$\lim _{x \rightarrow 0} \frac{x^2 \sin ^2(3 x)+\sin ^4(6 x)}{(1-\cos 3 x)^2}=$

આપેલ લક્ષની કિંમત શોધો: $\mathop {\lim }\limits_{z \to 1} \frac{z^{1/3}-1}{z^{1/6}-1}$

$\lim _{x \rightarrow 1} \left( \frac{x+x^2+x^3+\ldots+x^n-n}{x-1} \right) = $

$\lim _{x \rightarrow \infty} x\left(\log \left(1+\frac{x}{2}\right)-\log \frac{x}{2}\right) = $

દરેક $t \in \mathbb{R}$ માટે,ધારો કે $[t]$ એ $t$ થી નાનો અથવા તેના જેટલો મહત્તમ પૂર્ણાંક છે. તો $\lim_{x \to 1^+} \frac{(1 - |x| + \sin |1 - x|) \sin (\frac{\pi}{2} [1 - x])}{|1 - x|^2}$ ની કિંમત શોધો.

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