જો $A \neq 0$ અને $x > 0$ હોય,તો $\lim _{n \rightarrow \infty} \frac{\cos x - e^{nx}}{1 - A e^{nx}} = $

$\mathop {\lim }\limits_{n \to \infty } \left( \frac{1}{2} + \frac{1}{{{2^2}}} + \frac{1}{{{2^3}}} + ... + \frac{1}{{{2^n}}} \right)$ ની કિંમત શોધો.

$\mathop {\lim}\limits_{x \to 1} \left[ {\left[ {\frac{4}{{{x^2} - {x^{ - 1}}}} - \frac{{1 - 3x + {x^2}}}{{1 - {x^3}}}} \right]^{ - 1} + \frac{{3 \cdot ({x^4} - 1)}}{{{x^3} - {x^{ - 1}}}}} \right] = $

$\lim _{x \rightarrow 8} \frac{\sqrt{1+\sqrt{1+x}}-2}{x-8}$ ની કિંમત શોધો.

$\lim _{x \rightarrow 0} \frac{1-\cos (1-\cos x)}{\sin ^4 x} = $

જો $\lim_{n \rightarrow \infty} \frac{(n+1)^{k-1}}{n^{k+1}}[(nk+1)+(nk+2)+\ldots+(nk+n)] = 33 \cdot \lim_{n \rightarrow \infty} \frac{1}{n^{k+1}} \cdot [1^k + 2^k + 3^k + \ldots + n^k]$ હોય,તો $k$ ની પૂર્ણાંક કિંમત $....$ છે.