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I have this:

$$-\frac{x}{e^{\frac{x}{10}}}-\frac{10}{e^{\frac{x}{10}}}$$

Do I need to derivate the numerator and denominator of all this expression or just :

$$-\frac{x}{e^{\frac{x}{10}}}$$

And

$$-\frac{10}{e^{\frac{x}{10}}}$$

Keeps the same?

Thanks.

Simon Fraser
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Limanido
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    If you apply l'Hopital, you must use it according to what the theorem says, not according to what you think it does. –  Jan 02 '20 at 14:54
  • But you wouldn't apply the Hospital in this example anyway. The numerator has polynomial growth, while the denominator has exponential growth. So you just write down the answer: $0$. – Angina Seng Jan 02 '20 at 14:59
  • (However, as long as you apply it according to what the theorem says, you can do anything) –  Jan 02 '20 at 14:59
  • Also, the second fraction is not an indeterminate form. Its limit is simply $0$. – bjorn93 Jan 02 '20 at 14:59

2 Answers2

2

There are several ways to solve this. Here is one:

$$\begin{align*}\lim_{x \to \infty} \left( -\dfrac{x}{ e^{ \tfrac{x}{10} } } - \dfrac{ 10 }{ e^{ \tfrac{x}{10} } } \right) & = \lim_{ x \to \infty} -\dfrac{x+10}{e^{\tfrac{x}{10}}} \\ & \underset{\infty}{\overset{\infty}{=}} -\lim_{x \to \infty} \dfrac{1}{\dfrac{1}{10}\cdot e^{\tfrac{x}{10}}} = 0\end{align*}$$

SlipEternal
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Note that as $x\to \infty$ the second fraction goes to zero and the L'Hospital's rule does not apply to it.

Thus you just need to work on the first fraction.

Mohammad Riazi-Kermani
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