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Question

$$\displaystyle \lim_{n \rightarrow\infty}\displaystyle \frac{2.3^{\mathrm{n}+1}-3.5^{\mathrm{n}+1}}{2.3^{\mathrm{n}}+3.5^{\mathrm{n}}}=$$


A
5
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B
15
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C
5
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D
0
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Solution

The correct option is C $$-5$$
$$\displaystyle \lim_{n \rightarrow\infty}\displaystyle \frac{2.3^{\mathrm{n}+1}-3.5^{\mathrm{n}+1}}{2.3^{\mathrm{n}}+3.5^{\mathrm{n}}}=\lim_{n \rightarrow\infty}\displaystyle \frac{6.3^{\mathrm{n}}-15.5^{\mathrm{n}}}{2.3^{\mathrm{n}}+3.5^{\mathrm{n}}}$$
$$\displaystyle =\lim_{n \rightarrow\infty}\displaystyle \frac{6.\left(\frac{3}{5}\right)^{\mathrm{n}}-15}{2.\left(\frac{3}{5}\right)^{\mathrm{n}}+3}$$, dividing Nr and Dr by $$5^n$$
$$=\displaystyle \frac{-15}{3}=-5$$, since  $$\dfrac{3}{5}<1$$

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