$\lim _{z \rightarrow 1} \frac{z^{1/3}-1}{z^{1/6}-1} = $

$\lim _{n \rightarrow \infty} \frac{\left[6^2+12^2+18^2+\ldots+(6 n)^2\right]^2}{[5+10+15+\ldots+5 n]\left[2^3+4^3+6^3+\ldots+(2 n)^3\right]} =$

$\mathop {\lim }\limits_{m \to \infty } {\left( {\cos \frac{x}{m}} \right)^m} = $

$\lim _{x \rightarrow \infty}\left(\frac{x+8}{x+1}\right)^{x+5} = \dots$

જો ${S_n} = \sum\limits_{k = 1}^n {{a_k}} $ અને $\mathop {\lim }\limits_{n \to \infty } {a_n} = a,$ હોય,તો $\mathop {\lim }\limits_{n \to \infty } \frac{{{S_{n + 1}} - {S_n}}}{{\sqrt {\sum\limits_{k = 1}^n k } }}$ ની કિંમત શોધો.

$\mathop {\lim }\limits_{x \to 0} \frac{{{{(1 + x)}^{1/x}} - e + \frac{1}{2}ex}}{{{x^2}}}$ ની કિંમત શોધો.