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

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(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

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