જો $\lim_{x \to \infty} \frac{(2x - 1)^{19} \cdot (3x + 2)^{11}}{(6x - 5)^{30}} = 2^a \cdot 3^b$ હોય, તો $a + b = $

  • A
    $-30$
  • B
    $-11$
  • C
    $-19$
  • D
    $30$

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ધારો કે $[.]$ એ મહત્તમ પૂર્ણાંક વિધેય દર્શાવે છે. વિધાન $(A) : \lim_{x \rightarrow \infty} \frac{[x]}{x} = 1$. કારણ $(R) : f(x) = x - 1, g(x) = [x], h(x) = x$ અને $\lim_{x \rightarrow \infty} \frac{f(x)}{x} = \lim_{x \rightarrow \infty} \frac{h(x)}{x} = 1$.

$\mathop {\lim }\limits_{x \to 1} \frac{{{x^3} - 1}}{{{x^2} + 5x - 6}} = $

$\lim _{n}$ ${\rightarrow \infty} \sqrt{2} \left[ \frac{(2+\sqrt{2})^n + (2-\sqrt{2})^n}{(2+\sqrt{2})^n - (2-\sqrt{2})^n} \right] =$

જો $l = \lim_{x \rightarrow 0} \frac{x}{|x| + x^2}$ હોય,તો $l$ ની કિંમત શું છે?

જો $\lim _{x \rightarrow \infty}\left(\left(\frac{e}{1-e}\right)\left(\frac{1}{e}-\frac{x}{1+x}\right)\right)^x=\alpha$ હોય,તો $\frac{\log _e \alpha}{1+\log _e \alpha}$ ની કિંમત શોધો:

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