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Mathematics Question on limits and derivatives

lim⁡n→∞   2n+1+3n+12n+3n\underset{n\to \infty }{\mathop{\lim }}\,\,\,\frac{{{2}^{n+1}}+{{3}^{n+1}}}{{{2}^{n}}+{{3}^{n}}} is equal to

A

00

B

11

C

22

D

33

Answer

33

Explanation

Solution

lim⁡n→∞   2n+1+3n+12n+3n\underset{n\to \infty }{\mathop{\lim }}\,\,\,\frac{{{2}^{n+1}}+{{3}^{n+1}}}{{{2}^{n}}+{{3}^{n}}}
lim⁡n→∞    2.2n+3.3n2n+3n\underset{n\to \infty }{\mathop{\lim }}\,\,\,\,\frac{{{2.2}^{n}}+{{3.3}^{n}}}{{{2}^{n}}+{{3}^{n}}}
=lim⁡n→∞   2.(23)n+3(23)n+1=\underset{n\to \infty }{\mathop{\lim }}\,\,\,\frac{2.{{\left( \frac{2}{3} \right)}^{n}}+3}{{{\left( \frac{2}{3} \right)}^{n}}+1}
=0+30+1=3=\frac{0+3}{0+1}=3