$\mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {1 + x} - \sqrt {1 - x} }}{{{{\sin }^{ - 1}}x}} = $

  • A
    $2$
  • B
    $1$
  • C
    $-1$
  • D
    આમાંથી કોઈ નહીં

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ધારો કે $f(x)$ એ $x = h$ આગળ વિકલનીય છે. તો $\lim_{x \to h} \frac{(x + h)f(x) - 2hf(h)}{x - h}$ ની કિંમત શોધો.

$\mathop {\lim }\limits_{x \to 1 } \frac{{\left( {\log \left( {1 + x} \right) - \log 2} \right)\left( {3 \cdot 4^{x - 1} - 3x} \right)}}{{\left( {{{\left( {7 + x} \right)}^{1/3}} - {{\left( {1 + 3x} \right)}^{1/2}}} \right)\sin \pi x}}$ ની કિંમત શોધો.

લક્ષની કિંમત શોધો: $\mathop {\lim }\limits_{x \to 0} \frac{{\int\limits_0^x (\tan^{-1} t)^2 dt}}{{\sin x - x}}$

જો $a > 0$ અને $\lim _{x \rightarrow a} \frac{a^x - x^a}{x^x - a^a} = -1$ હોય,તો $a$ ની કિંમત શોધો.

$\mathop {\lim }\limits_{x \to 0} {\left\{ {\tan \left( {\frac{\pi }{4} + x} \right)} \right\}^{1/x}} = $

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