limits for positive and negative infinity

Question

It says we use the l'hospital's rule, however I don't understand because the limit for positive infinity and negative infinity are different. Please help!

Answer 1

First we find $$\lim_{x\to-\infty}\frac{5+2^x}{1-2^x}.$$ But $2^x\to 0$ as $x\to-\infty$, and therefore the required limit is $\frac{5}{1}$.

Next we find $$\lim_{x\to\infty}\frac{5+2^x}{1-2^x}.$$ Divide top and bottom by $2^x$. So we want $$\lim_{x\to\infty}\frac{5\cdot 2^{-x}+1}{e^{-x}-1}.$$ But $2^{-x}\to 0$ as $x\to\infty$, and therefore the required limit is $\frac{1}{-1}$.

Remark: For the limit as $x\to -\infty$, we cannot use L'Hospital's Rule directly, since top and bottom both have finite limits. We could after some manipulation rewrite things so that L'Hospital's Rule can be used, but it is harder than the direct approach we took.

For the limit as $x\to\infty$, we can use L'Hospital's Rule. Recall that the derivative of $2^x$ is $(\ln 2)2^x$. L'Hospital's Rule then gives $$\lim_{x\to\infty}\frac{5+2^x}{1-2^x}=\lim_{x\to\infty} \frac{(\ln 2)2^x}{-(\ln 2)2^x}.$$ Cancellation now gives that the limit is $-1$. I much prefer the approach of the main answer above.