# Finding limits using l'Hospital's Rule

Find the limit. Use l'Hospital's Rule if appropriate. If there is a more elementary method, consider using it.

Find

$$\lim_{x\to \infty} x^7\cdot(e^ {-x})^6$$

My first instinct would be to take the natural log, and if I remember my log rules correctly that would be $$7\cdot\ln(x)-6\cdot\ln\left(e^{-x}\right)$$

I'm not sure if you can bring down the $-x$ as well. I know $\ln(e)=1$, so I think it would then be $$7\cdot\ln(x)-6\cdot(-x)$$

I'm not sure where to go from here and how you rewrite this to be able to use l'Hospital's Rule. Can anyone help?

• I don't think there's anything simpler than L'Hopital, unless you know that $\ln(x)/x^r\rightarrow 0$ as $x\rightarrow\infty$ for instance. – Olivier Moschetta Oct 10 '16 at 20:49
• @OlivierMoschetta Your statement itself can be shown using l'Hospital's rule. – Fan Zheng Oct 10 '16 at 20:50
• you should have $+6(-x)$ instead of $-6(-x)$. – hamam_Abdallah Oct 10 '16 at 20:51
• Pedagogically, if your instructor is making you use L'hospital's rule for this, he or she is not being very wise. Using L'hospital's rule for this doesn't teach students anything. It should be immediately apparent that the exponential in the denominator dominates the polynomial in the numerator. – MathematicsStudent1122 Oct 10 '16 at 20:52

rewrite the original as $\frac{x^7}{e^{x^6}}$; then you could apply L'Hopital's rule, as it is of the type '$\frac{\infty}{\infty}$'. It should be clear that the exponential function grows much faster than the polynomial and hence the limit goes to 0.
You can write the function as :$$y=\frac{x^7}{e^{6x}}$$
Write the function as :$$y=\frac{x^7}{e^{6x}}$$ and apply L'Hopital law until the numerator becomes a constant and remember $e^x$ is almost equal to $0$ provided $x<0$ and $|x|$is a large no.