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My objective is to solve the following limit $$\lim_{x \to \infty} \frac{a^x - b^x}{a^{x+1} - b^{x+1}},$$ where $a$ and $b$ are real constants, such that $a \gt b \gt 0$.

I (apparently) managed to solve it in two different ways, leading to two distinct answers, which of course, means something is wrong.

Method 1: $$\lim_{x \to \infty} \frac{a^x - b^x}{a^{x+1} - b^{x+1}}\frac{a^{-x}}{a^{-x}}=\lim_{x \to \infty} \frac{1 - (\frac{b}{a})^x}{a - a(\frac{b}{a})^{x+1}}$$ using the L´Hospital rule, we get $$\lim_{x \to \infty} \frac{-(\log{\frac{b}{a}})(\frac{b}{a})^x}{-a(\log{\frac{b}{a}})(\frac{b}{a})^{x+1}} = \frac{1}{a}(\frac{b}{a})^{-1} = \frac{1}{b}.$$

Method 2: $$\lim_{x \to \infty} \frac{a^x - b^x}{a^{x+1} - b^{x+1}}\frac{b^{-x}}{b^{-x}}=\lim_{x \to \infty} \frac{(\frac{a}{b})^x - 1}{b(\frac{a}{b})^{x+1} - b}$$ using the L´Hospital rule, we get $$\lim_{x \to \infty} \frac{(\log{\frac{a}{b}})(\frac{a}{b})^x}{b(\log{\frac{a}{b}})(\frac{a}{b})^{x+1}} = \frac{1}{b}(\frac{a}{b})^{-1} = \frac{1}{a}.$$

What am I doing wrong?

Edit: added constraints for $a$ and $b$.

durdi
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1 Answers1

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Note that after this step

$$\lim_{x \to \infty} \frac{a^x - b^x}{a^{x+1} - b^{x+1}}\frac{a^{-x}}{a^{-x}}=\lim_{x \to \infty} \frac{1 - (\frac{b}{a})^x}{a - a(\frac{b}{a})^{x+1}}$$

we need some condition to apply l'Hopital (i.e. indeterminate form $\frac{\infty}{\infty}$ or $\frac{0}{0}$) and similarly for Method 2. That's not the correct way to solve the limit.

To solve note that for $x$ real we need $a,b>0$ and we can distinguish the two cases $a>b$ or $b>a$. For $x$ integer we can also consider $a,b <0$.

Notably for $x$ real and $a,b>0$

  • for $a>b\implies \lim_{x \to \infty} \frac{a^x - b^x}{a^{x+1} - b^{x+1}}\frac{a^{-x}}{a^{-x}}=\lim_{x \to \infty} \frac{1 - (\frac{b}{a})^x}{a - a(\frac{b}{a})^{x+1}}=\frac1a$

  • for $b>a\implies \lim_{x \to \infty} \frac{a^x - b^x}{a^{x+1} - b^{x+1}}\frac{b^{-x}}{b^{-x}}=\lim_{x \to \infty} \frac{(\frac{a}{b})^x - 1}{b(\frac{a}{b})^{x+1} - b}=\frac1b$

For $x$ integer with $a\cdot b<0$

  • for $|a|>|b|\implies \lim_{x \to \infty} \frac{a^x - b^x}{a^{x+1} - b^{x+1}}=\lim_{x \to \infty} \frac{a^{x}}{a^{x+1}}\frac{1 - (\frac{b}{a})^x}{1 - (\frac{b}{a})^{x+1}}=\frac1a$

  • for $|b|>|a|\implies \lim_{x \to \infty} \frac{a^x - b^x}{a^{x+1} - b^{x+1}}=\lim_{x \to \infty} \frac{b^{x}}{b^{x+1}}\frac{(\frac{a}{b})^x - 1}{(\frac{a}{b})^{x+1} - 1}=\frac1b$

user
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  • Had you originally answered the question fully, I would not have had to dv, and then, after your corrective edit, un-downvote. I can't always make it back to recheck poor answers I've felt I've had to downvote, so consider yourself lucky that your -2 becomes a -1 now. – amWhy May 09 '18 at 20:11
  • @amWhyNote that the OP was asking for "What am I doing wrong?" then in the first version I tried to answer to that precisely question. Then, after downvotes, I was forced to add some more information. I had the doubt the the first considerations were wrong or misleading. Anyway thanks for your comment for the dv and uv. Bye – user May 09 '18 at 20:16
  • True...well, for the sake of this site and down the "road" with new users, your answer will surely be helpful. Take care. – amWhy May 09 '18 at 20:34