lim _{n rightarrow infty} frac{3 cdot 2^{n+1} | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) To evaluate the limit $\lim _{n \rightarrow \infty} \frac{3 \cdot 2^{n+1}-4 \cdot 5^{n+1}}{5 \cdot 2^{n}+7 \cdot 5^{n}}$,we divide the numerator a

Vedclass · Accepted Answer

$-\frac{20}{7}$

Vedclass · Answer

(C) To evaluate the limit $\lim _{n \rightarrow \infty} \frac{3 \cdot 2^{n+1}-4 \cdot 5^{n+1}}{5 \cdot 2^{n}+7 \cdot 5^{n}}$,we divide the numerator and the denominator by $5^n$: $\lim _{n \rightarrow \infty} \frac{3 \cdot 2 \cdot 2^{n} - 4 \cdot 5 \cdot 5^{n}}{5 \cdot 2^{n} + 7 \cdot 5^{n}}$ $= \lim _{n \rightarrow \infty} \frac{6 \cdot 2^{n} - 20 \cdot 5^{n}}{5 \cdot 2^{n} + 7 \cdot 5^{n}}$ Dividing by $5^n$: $= \lim _{n \rightarrow \infty} \frac{6 \cdot (\frac{2}{5})^{n} - 20}{5 \cdot (\frac{2}{5})^{n} + 7}$ Since $\lim _{n \rightarrow \infty} (\frac{2}{5})^{n} = 0$,we get: $= \frac{6 \cdot 0 - 20}{5 \cdot 0 + 7} = -\frac{20}{7}$

$\lim _{n \rightarrow \infty} \frac{3 \cdot 2^{n+1}-4 \cdot 5^{n+1}}{5 \cdot 2^{n}+7 \cdot 5^{n}}$ is equal to

Similar Questions

$\mathop {\lim }\limits_{x \to 0} \frac{{x({e^x} - 1)}}{{1 - \cos x}} = $

The value of $\lim\limits_{n \rightarrow \infty} 6 \tan \left\{\sum\limits_{r=1}^{n} \tan ^{-1}\left(\frac{1}{r^{2}+3 r+3}\right)\right\}$ is equal to

The limit of the perimeter of the regular $n$-gons inscribed in a circle of radius $R$ as $n \to \infty$ is

Let $f(x) = \frac{\sqrt{x+3}}{x+1}$. Then the value of $\lim_{x \rightarrow -3^{-}} f(x)$ is

$\mathop {\lim }\limits_{n \to \infty } \left[ {\frac{{{1^3} + {2^3} + {3^3} + \dots + {n^3}}}{{{n^4}}}} \right] = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module