3

There is a problem from my textbook: Let $a$ be a constant with $a>3$ and $P$ be the intersection point between two curves $y=a^{x-1}$ and $y=3^x$. We denote the $x$-coordinate of $P$ by $k$

Then, compute $$ \lim_{n\to\infty}\left(\frac{(\frac{a}{3})^{n+k}}{(\frac{a}{3})^{n+1}+1}\right)$$

My approach is : $$ a^{x-1}=3^x$$ $$ (\frac{a}{3})^x=a$$ As $x$ here equals to $k$ so $$(\frac{a}{3})^k=a$$ Substituting back to the original equation $$ \lim_{n\to\infty}\left(\frac{a(\frac{a}{3})^{n}}{(\frac{a}{3})(\frac{a}{3})^{n}+1}\right)$$ AS the degrees of the leading terms are equal we should take the quotient of the coefficients of them $$\frac{a}{\frac{a}{3}}$$ So answer is 3

Can someone look through and judge weather this way is correct becouse I don`t have the right answer?

Islombek Rakhmatov
  • 81
  • It looks perfectly correct to me. Why do you think your answer isn't right? One note, though: the "degrees are equal" logic is only valid here because $\frac{a}{3}>1$. – zipirovich Jul 16 '17 at 19:05
  • @zipirovich what if a/3 is not greater than 1 – Islombek Rakhmatov Jul 16 '17 at 19:34
  • Here we know that $\frac{a}{3}>1$ because it's given that $a>3$. This condition is important because $b^n\to\infty$ as $n\to\infty$ (so that we can call it a "leading term") only when $b>1$. If $0<a<3$, then $0<\frac{a}{3}<1$, and $\left(\frac{a}{3}\right)^n\to0$ as $n\to\infty$; thus the limit will be equal to $0$. If $a=3$, then $\frac{a}{3}=1$ and $\left(\frac{a}{3}\right)^n=1$ for any $n$; the limit is then $\frac{3}{2}$. – zipirovich Jul 16 '17 at 19:45

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