# Evaluating $\lim_{x\to\infty} \frac{3-3^x}{5-5^x}$

I'm trying to get back up to speed on calculus 30 years after I last took a course on it, so I dug out my old undergraduate textbook and have been working through the problems on my own. I came across this problem in the section on L'Hopital's rule. Obviously, I need to apply L'Hopital's rule, but when I do, I end up with $$\lim_{x\to\infty} \frac{3^x \ln 3}{5^x \ln 5}$$, which I see as infinity, but the answer in the back of the book is $$3/5$$. Any hints on what I'm missing here?

• Divide numerator and denominator by $5^x$ of the original problem May 30, 2019 at 19:20
• "but the answer in the back of the book is $\frac{3}{5}$" This seems like you are either giving us incorrect information or the book got it wrong. Are you sure you are looking at the limit as $x$ approaches positive infinity rather than negative infinity? Alternatively, are you sure that these are meant to be exponents rather than multiplications? May 30, 2019 at 19:24
• A quick and dirty way to find limits like these is to completely ignore any and every term which is not the "fastest growing" compared to the others around it. $3^x$ in the numerator is much larger than $3$, same with $5^x$ compared to $5$, so you would have $\lim\limits_{x\to\infty} \frac{3-3^x}{5-5^x}$ should act the same as $\lim\limits_{x\to\infty}\frac{-3^x}{-5^x}$ which simplifies to $\lim\limits_{x\to\infty}\left(\frac{3}{5}\right)^x$ which is zero. This idea of "ignoring smaller growing functions" can be made rigorous, e.g. by performing the divisions as done in answers below. May 30, 2019 at 19:26
• You are correct, JMoravitz - I did read and transcribe it incorrectly, it was negative infinity. Although I'm still struggling to figure out the steps, at least the answer makes sense now, since the graph of the function converges to 3/5. May 31, 2019 at 21:01

Hint: Write your term in the form $$\frac{\frac{3}{5^x}-\left(\frac{3}{5}\right)^x}{\frac{5}{5^x}-1}$$
For $$x\rightarrow+\infty$$ we obtain: $$\frac{3-3^x}{5-5^x}=\frac{\frac{3}{5^x}-\left(\frac{3}{5}\right)^x}{\frac{5}{5^x}-1}\rightarrow0.$$ Also, by the Hospital's rule we obtain: $$\lim_{x\rightarrow+\infty}\frac{3-3^x}{5-5^x}=\lim_{x\rightarrow+\infty}\frac{-3^x\ln3}{-5^x\ln5}=0.$$
$$3,5$$ are infinitesimal, so it becomes:
$$\lim_{x\to \infty} \left(\frac 35\right)^x=0$$