Question

find the limit. use lhospitals rule where appropriate. if there is a more elementary method, consider using it.
\\( \lim _ { x \
ightarrow \infty } x ^ { 7 } e ^ { - x ^ { 6 } } \\)

Explanation:

Step1: Rewrite the limit

Rewrite \( \lim_{x
ightarrow\infty}x^{7}e^{-x^{6}} \) as \( \lim_{x
ightarrow\infty}\frac{x^{7}}{e^{x^{6}}} \). As \( x
ightarrow\infty \), we have the indeterminate form \( \frac{\infty}{\infty} \).

Step2: Apply L'Hospital's Rule

Differentiate the numerator and denominator. The derivative of \( y = x^{7} \) is \( y^\prime=7x^{6} \), and the derivative of \( y = e^{x^{6}} \) using the chain - rule (\( (e^{u})^\prime=e^{u}\cdot u^\prime \), where \( u = x^{6} \), \( u^\prime = 6x^{5} \)) is \( y^\prime=e^{x^{6}}\cdot6x^{5} \). So, \( \lim_{x
ightarrow\infty}\frac{x^{7}}{e^{x^{6}}}=\lim_{x
ightarrow\infty}\frac{7x^{6}}{6x^{5}e^{x^{6}}}=\lim_{x
ightarrow\infty}\frac{7x}{6e^{x^{6}}} \).

Step3: Analyze the new limit

As \( x
ightarrow\infty \), \( e^{x^{6}}\) grows much faster than \( x \). We can also think of it as applying L'Hospital's Rule again. If we consider \( \lim_{x
ightarrow\infty}\frac{7x}{6e^{x^{6}}} \), and rewrite it as \( \frac{7}{6}\lim_{x
ightarrow\infty}\frac{x}{e^{x^{6}}} \). Let \( t=x^{6} \), then \( x = t^{\frac{1}{6}} \) and as \( x
ightarrow\infty \), \( t
ightarrow\infty \). So, \( \frac{7}{6}\lim_{x
ightarrow\infty}\frac{x}{e^{x^{6}}}=\frac{7}{6}\lim_{t
ightarrow\infty}\frac{t^{\frac{1}{6}}}{e^{t}} \).
We know that for any positive real number \( a \), \( \lim_{t
ightarrow\infty}\frac{t^{a}}{e^{t}} = 0 \) (by repeatedly applying L'Hospital's Rule \( \lceil a
ceil+1 \) times or using the fact that the exponential function \( y = e^{t} \) has a faster growth rate than any polynomial function \( y=t^{a}\)).

Answer:

\( 0 \)