$\lim_{x \to \infty} e^x - \frac{e^x}{x+1}$ Application of L'Hopital's Rule

Question

I want to know if I can "use" a limit after I've used L'hopital's rule on it? I'm not sure how to better word it, but I can show you what I tried, maybe you can tell me if it is right or why it is wrong. $$\lim_{x \to \infty} e^x - \frac{e^x}{x+1}$$ We can split this into two limits $$\lim_{x \to \infty} e^x - \lim_{x \to \infty} \frac{e^x}{x+1}$$ Now since the limit on the right side is infinity over infinity, we can apply L'Hopital's rule

$$\lim_{x \to \infty} e^x - \lim_{x \to \infty} \frac{e^x}{1}$$ Now we can join the two limits back (I am "reusing" the limit after applying L'hopital...is this allowed?) $$\lim_{x \to \infty} e^x - e^x$$ Subtracting we have $$\lim_{x \to \infty} 0 = 0$$

Answer 1

No we are not allowed to do that, we need to proceed as follows

$$\lim_{x \to \infty} e^x - \frac{e^x}{x+1}=\lim_{x \to \infty} \frac{xe^x}{x+1}$$

and then apply l'Hopital since the expression is in the form $\frac{\infty}{\infty}$.

Recall indeed that we can apply l'Hopital$^{(*)}$ for expressions $\frac{f(x)}{g(x)}$ in the form $\frac{\infty}{\infty}$ or $\frac{0}{0}$ or also, as an extension, for the case $f(x)^{g(x)}=e^{g(x)\log(f(x))}$ when $g(x)\log(f(x))$ is in the indeterminate form $\frac{\infty}{\infty}$ or $\frac{0}{0}$.

$(*)$Note

As noticed by Mark Viola, it is not necessary that the numerator approaches $\infty$ to apply l'Hopital, indeed the numerator need not even have a limit provided that the other conditions hold and the denominator approaches $\infty$ (reference wiki article).

Answer 2

The theorem is $\lim_{x \to a} \ (f(x) + g(x)) = \lim_{x \to a} \ f(x) + \lim_{x \to a} \ g(x)$ is valid in general, if both limits $\lim_{x \to a}\ f(x) \ \text{and} \lim_{x \to a}\ g(x)$ are individually finitely exists.

Answer 3

Alternatively, you can pull $e^x$ out of brackets and avoid using L'Hospital's rule: $$\lim_{x \to \infty} e^x - \frac{e^x}{x+1}=\lim_{x \to \infty} e^x\left(1 - \frac{1}{x+1}\right)=\infty\cdot (1-0)=\infty\cdot 1=\infty.$$ For practice: 1) $\lim_\limits{x\to\infty} (x^2-x)$; 2) $\lim_\limits{x\to\infty} (x-\ln x)$; 3) $\lim_\limits{x\to\infty} (e^x-x\ln x)$.

Extra reading on MSE: 1, 2, 3, .