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

લક્ષ શોધો: $\mathop {\lim }\limits_{x \to 1} (x^3 - x^2 + 1)$

કિંમત શોધો: $\mathop {\lim }\limits_{x \to 0} \frac{\sqrt{1+x}-1}{x}$

જો $n$ એક પૂર્ણાંક હોય,તો $\mathop {\lim }\limits_{x \to n + 0} (x - [x]) = $

જો $\alpha=\lim _{x \rightarrow 0} \frac{x \cdot 2^x-x}{1-\cos x}$ અને $\beta=\lim _{x \rightarrow 0} \frac{x \cdot 2^x-x}{\sqrt{1+x^2}-\sqrt{1-x^2}}$ હોય,તો

સીમાની ગણતરી કરો: $\lim _{x}$ ${\rightarrow \infty} \frac{(\sqrt{3 x+1}+\sqrt{3 x-1})^6+(\sqrt{3 x+1}-\sqrt{3 x-1})^6}{\left(x+\sqrt{x^2-1}\right)^6+\left(x-\sqrt{x^2-1}\right)^6} x^3$