Question

Evaluate ${ \displaystyle \lim_{ \bf n \to \infty} \bf \dfrac{{3}^{n+1} + {4}^{n+1}}{3^{n} + 4^{n}} }$

Can it be solved by L'hopital rule ? ​

Answer 1

${ \displaystyle \lim_{ \bf n \to \infty} \bf \dfrac{{3}^{n+1} + {4}^{n+1}}{3^{n} + 4^{n}} =4}$

Answer 2

Answer:

The answer to this question is 4

Step-by-step explanation:

We know that,

$\lim_{n \to \infty} (\frac{a}{b})^n = 0~\text{if}~ a<b$

${ \displaystyle \lim_{ \bf n \to \infty} \bf \frac{{3}^{n+1} + {4}^{n+1}}{3^{n} + 4^{n}}$

$= \lim_{n \to \infty} \frac{4^{n+1}(\frac{3^{n+1}}{4^{n+1}}+1)}{4^{n}(\frac{3^{n}}{4^{n}}+1)}$

$= 4 \lim_{n \to \infty} \frac{(\frac{3}{4})^{n+1} +1}{(\frac{3}{4})^{n} +1}\\= 4 (\frac{0+1}{0+1})\\= 4$

and yes, this can also be solved using L'hospital's rule