$\mathop {\lim }\limits_{x \to 0} x \log (\sin x) = $

  • A
    $0$
  • B
    $-\infty$
  • C
    $1$
  • D
    $None \ \text{of these}$

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यदि $\log (1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\ldots \infty$ और $\lim _{x \rightarrow 0} \frac{\log (1+x)^{1+x}}{x^2}-\frac{1}{x}=k$ है,तो $12 k=$

सीमा का मूल्यांकन करें: $\lim _{x \rightarrow 1} \left[\frac{\sqrt{x}-1}{\log x}\right]$

$\mathop {\lim }\limits_{x \to 0} \frac{{{{27}^x} - {9^x} - {3^x} + 1}}{{\sqrt 5 - \sqrt {4 + \cos x} }}$ का मान है

$\mathop {\lim }\limits_{x \to 0^+} {x^x} = $

मान लीजिए $[x]$ $x$ से कम या उसके बराबर सबसे बड़ा पूर्णांक दर्शाता है और $k \geq 2$ एक पूर्णांक है। तो $\lim_{x \rightarrow k} \frac{\sin \left(2 \pi\left([x]-\left[\frac{x}{k}\right]\right)-x\right)+\sin k}{x-k} = $

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