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

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 $$\lim_{x \to \infty} 0 = 0$$

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).

• It is not necessary that the numerator approaches $\infty$ to apply LHR. In fact, the numerator need not even have a limit provided that the other conditions hold and the denominator approaches $\infty$. So please, modify your answer to correct the misstatement. – Mark Viola Nov 28 '18 at 23:50
• @MarkViola Sorry I'm not used with this generalized version, could you please give me a reference to it on order to correct my statement. I hope the other parts are ok. Thanks – gimusi Nov 28 '18 at 23:55
• Just to clarify, there is no case where it would be okay to reincorporate a limit upon which I have applied L'Hopital's rule into the general limit statement? It is not clear for me whether my technique is incorrect due to that or some other factor. – Danielle Nov 28 '18 at 23:56
• @MarkViola I've found that related OP – gimusi Nov 28 '18 at 23:57
• – Mark Viola Nov 28 '18 at 23:57

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.

• Correct, but then can I apply L'Hopital's rule to one of the limits and then reincorporate that limit back into the general limit. – Danielle Nov 29 '18 at 0:00
• The important part in prashant's comment is "if both limits exists and are finite (!)". And in your case they were both infinite, so you cannot "reincorporate", that is apply the sum rule. One way to see that your original calculation is wrong is to note your result: zero. The correct answer is infinity... But prashant's comment explains why it is wrong. – Antoshka Nov 29 '18 at 1:04

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, .