Question

The value of $\underset{n\to \infty }{lim}{\left({\left(1.5\right)}^n+{\left[{\left(1+0.0001\right)}^{10000}\right]}^n\right)}^{\frac{1}{n}}$, where [.] denotes the greatest integer function is

• A. $1$
• B. $\frac{1}{2}$
• C. $Does{n}^{\prime }texist$
• D. $2$

Answer 1

The correct option is D $2$

$\begin{array}{c}\underset{n\to \infty }{lim}{\left({\left(1.5\right)}^n+{\left[{\left(1+0.0001\right)}^{10000}\right]}^n\right)}^{\frac{1}{n}}\\ \underset{n\to \infty }{lim}{\left({\left(1.5\right)}^n+{\left[{}^{10000}{C}_0\left(1\right){+}^{10000}{C}_1\left(0.0001\right){+}^{10000}{C}_2{\left(0.0001\right)}^2+......\right]}^n\right)}^{\frac{1}{n}}\\ \underset{n\to \infty }{lim}{\left({\left(1.5\right)}^n+{\left[1+\left(10000\times 0.0001\right)+decimal\_nos\right]}^n\right)}^{\frac{1}{n}}\\ \underset{n\to \infty }{lim}{\left({\left(1.5\right)}^n+{\left[1+1+0\left(approx\right).....\right]}^n\right)}^{\frac{1}{n}}\\ \underset{n\to \infty }{lim}{\left({\left(1.5\right)}^n+2^n\right)}^{\frac{1}{n}}=\underset{n\to \infty }{lim}{\left(2^n\left[{\left(\frac{1.5}{2}\right)}^n+1\right]\right)}^{\frac{1}{n}}\\ \underset{n\to \infty }{lim}{2}^{\frac{n}{n}}{\left({\left(\frac{3}{4}\right)}^n+1\right)}^{\frac{1}{n}}=2(1{)}^0=2\end{array}$