Question

If $0<p<q,$ then $\underset{n\to \infty }{\lim }{\left(q^n+p^n\right)}^{\frac{1}{n}}=?$

• A. $e$
• B. $p$
• C. $q$
• D. $0$

Answer 1

The correct option is C

$q$

Explanation for correct option:

Step-1: Simplify the given data.

Given, $\underset{n\to \infty }{\lim }{\left(q^n+p^n\right)}^{\frac{1}{n}};0<p<q$

$=\underset{n\to \infty }{\lim }{\left(q^n\left(1+\frac{p^n}{q^n}\right)\right)}^{\frac{1}{n}};0<p<q$

$=\underset{n\to \infty }{\lim }{\left({\left(q^n\right)}^{\frac{1}{n}}{\left\{1+{\left(\frac{p}{q}\right)}^n\right\}}^{{\left(\frac{q}{p}\right)}^n}\right)}^{\frac{1}{n}{\left(\frac{p}{q}\right)}^n};0<p<q$

Step-2: Apply formula: $\underset{\mathbf{x}\mathbf{\to }\mathbf{\infty }}{\mathbf{l}\mathbf{i}\mathbf{m}}{\mathbf{\left(}\mathbf{f}\left(x\right)\mathbf{\right)}}^{\left(g\left(x\right)\right)}\mathbf{=}{\mathit{e}}^{\underset{\mathbf{x}\mathbf{\to }\mathbf{\infty }}{\mathbf{l}\mathbf{i}\mathbf{m}}\mathbf{\left(}\mathbf{g}\left(x\right)\mathbf{\right)}\mathbf{[}\mathbf{f}\left(x\right)\mathbf{-}\mathbf{1}\mathbf{]}}$

$=q{e}^{\underset{n\to \infty }{\lim }\frac{1}{n}\left(1+{\left(\frac{p}{q}\right)}^n-1\right)},0<p<q$

$=q{e}^0,0<p<q$ $\left[\mathbf{\because }\frac{\mathbf{1}}{\mathbf{\infty }}\mathbf{=}\mathbf{0}\right]$

$=q$

Hence, correct answer is option $\left(C\right)$