Finding limits using l’Hôpital’s rule

Question

$\displaystyle\lim_{x\to \infty}(e^{3x} - 5x)^{1/x}$

$\displaystyle\ln y=\frac{\ln(e^{3x} - 5x)}{x}$

Now applying l’Hôpital’s rule three times I can get the answer $3$.

That means,

$\displaystyle\lim_{x\to\infty}(e^{3x} - 5x)^{1/x}=e^3$

But after finding this answer I understood that when I’m using l’Hôpital’s rule first time there isn't a $\frac{\infty}{\infty}$ form! There is an indeterminate form inside the $\ln$ which is $\ln (\infty-\infty)$ because $\lim_{x\to\infty}$

So which is the correct way to find the limits?

Answer 1

Indeed, $e^{3x}-5x$ is an indeterminate form $\infty-\infty$, but this doesn't mean you can do nothing about it. Just consider $$ \lim_{x\to\infty}(e^{3x}-5x)= \lim_{x\to\infty}e^{3x}\left(1-\frac{5x}{e^{3x}}\right)=\infty $$ because the part in parentheses has limit $1$. Thus you indeed have $$ \lim_{x\to\infty}\frac{\ln(e^{3x}-5x)}{x} $$ in the indeterminate form $\infty/\infty$ and you can apply l’Hôpital: $$ \lim_{x\to\infty}\frac{\ln(e^{3x}-5x)}{x}= \lim_{x\to\infty}\frac{3e^{3x}-5}{e^{3x}-5x}= \lim_{x\to\infty}\frac{e^{3x}\left(3-\dfrac{5}{e^{3x}}\right)}{e^{3x}\left(1-\dfrac{5x}{e^{3x}}\right)}=3 $$ (no need to go three times with l’Hôpital). The key fact used a few times here is that $$ \lim_{t\to\infty}\frac{t^k}{e^t}=0 $$ for every $k$.

Answer 2

Since $\exp(3x)-5x \to \infty$, we have $\ln (\exp(3x)-5x) \to \infty$, hence we do have the indeterminate form.

Remark:

Note the statement of L'hopital for $\lim_{x \to c}\frac{f_1(x)-f_2(x)}{g(x)}$ works when $$\lim_{x \to c} [f_1(x) - f_2(x)]=\infty=\lim_{x \to c}g(x)$$

Answer 3

Note that by standard limits

$$\frac{5x}{e^{3x}}\to 0$$

then

$$\frac{\ln(e^{3x} - 5x)}{x}=\frac{\ln\left(1-\frac{5x}{e^{3x}}\right)+\ln e^{3x}}{x}= \frac{5}{e^{3x}}\frac{\ln\left(1-\frac{5x}{e^{3x}}\right)}{\frac{5x}{e^{3x}}} +\frac{\ln e^{3x}}{x}=\\= \frac{5}{e^{3x}}\frac{\ln\left(1-\frac{5x}{e^{3x}}\right)}{\frac{5x}{e^{3x}}} +3\to0\cdot1+ 3=3$$