$\mathop {\lim }\limits_{x \to a} \frac{{{x^2} - {a^2}}}{{x - a}} = $

  • A
    $4a$
  • B
    $1$
  • C
    $2a$
  • D
    $0$

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Similar Questions

ધારો કે $S$ એ તમામ $(\alpha, \beta) \in \mathbb{R} \times \mathbb{R}$ નો ગણ છે કે જેથી $\lim _{x \rightarrow \infty} \frac{\sin(x^2)(\log_e x)^\alpha \sin(1/x^2)}{x^{\alpha \beta}(\log_e(1+x))^\beta} = 0$ થાય. તો નીચેનામાંથી કયું (કયા) સાચું છે?

જો $f(x) = \frac{2}{x - 3}$,$g(x) = \frac{x - 3}{x + 4}$ અને $h(x) = - \frac{2(2x + 1)}{x^2 + x - 12}$ હોય,તો $\lim_{x \to 3} [f(x) + g(x) + h(x)]$ ની કિંમત શોધો.

વિધાન $(A)$: $\lim _{x \rightarrow 0} \frac{1}{x} = \infty$
કારણ $(R)$: જેમ $x$ ની કિંમત ઘટે છે,તેમ $\frac{1}{x}$ ની કિંમત વધે છે.

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

$\lim _{x \rightarrow 0} \frac{e^x-e^{\sin x}}{2(x-\sin x)}$

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