Lots Of Limitzzzzz//-

as\;x\rightarrow 0, \frac{sinx}{x}\rightarrow 1^{-1}\;and\;\frac{tanx}{x}\rightarrow1^{+}

so as\;x\rightarrow 0, \frac{nsinx}{x}\rightarrow n^{-}\;and\;\frac{tanx}{x}\rightarrow n^{+}

so

\lim_{x\rightarrow 0}\left[\frac{nsinx}{x} \right]=n-1

\lim_{x\rightarrow 0}\left[\frac{ntanx}{x} \right]=n

x(1+(\frac{x^2+1}{x^3})^{\frac{1}{3}})\\ \lim _{x\to \infty}(1+\frac{1}{3}(\frac{x^2+1}{x^2})-\frac{2}{9}(\frac{(x^2+1)^{2}}{x^5})+.....)=\frac{1}{3}

PS: in 6 , fogot the powers [3] [3]

Let S = \left(1 + \frac{1}{n} \right)^{\frac{1}{2}} + \left(1 + \frac{1}{n} \right)^{\frac{2}{3}}+...+\left(1 + \frac{1}{n} \right)^{\frac{n}{n+1}}-n

Then S < \left(1 + \frac{1}{n} \right) + \left(1 + \frac{1}{n} \right)+...+\left(1 + \frac{1}{n} \right)-n = 1

Also

S = \left[\left(1 + \frac{1}{n} \right)^{\frac{1}{2}}-1 \right ] + \left[\left(1 + \frac{1}{n} \right)^{\frac{2}{3}}-1 \right] +...+\left[\left(1 + \frac{1}{n} \right)^{\frac{n}{n+1}} - 1 \right]

> \frac{1}{n} \left(\frac{1}{2} + \frac{2}{3}+...+\frac{n}{n+1} \right)

= \frac{1}{n} \left(n - \frac{1}{2} -\frac{1}{3}-... -\frac{n}{n+1} \right)

= 1 - \frac{1}{n} \left(\frac{1}{2} +\frac{1}{3}+... +\frac{1}{n+1} \right)

> 1 - \frac{\ln (n+1) - \ln 2}{n} (using the inequality between the area under the curve y = 1/1+x between x = 1 and x = n+1, and the area of smaller rectangle)

Hence 1 - \frac{\ln (n+1) - \ln 2}{n}<S<1

By Sandwich Theorem, the limit is seen to be 1 (using \lim_{n \rightarrow \infty} \frac{\ln (n+1)}{n} = 0)

edit: should have been ln (n+2) instead of ln (n+1)

If you use stirling approximation, you will get 1/e

4th one- this looks simpler to me:
\lim_{x \rightarrow \infty} (x^3+x^2+1)^{\frac{1}{3}} - x = \lim_{x \rightarrow \infty} \frac{x^2+1}{x^2+(x^3+x^2+1)^{\frac{2}{3}} + x(x^3+x^2+1)^{\frac{1}{3}}}

\ \lim_{x \rightarrow \infty} \frac{1+\frac{1}{x^2}}{1+\left(1 + \frac{1}{x} + \frac{1}{x^2}\right)^{\frac{2}{3}} + \left(1+\frac{1}{x} + \frac{1}{x^2}\right)^{\frac{1}{3}}} = \frac{1}{3}