find the limit. use lhospitals rule where appropriate. if there is a more elementary method, consider using…

Explanation:

Step1: Rewrite the limit

Rewrite ( \lim_{x\rightarrow\infty}x^{7}e^{-x^{6}} ) as ( \lim_{x\rightarrow\infty}\frac{x^{7}}{e^{x^{6}}} ). As ( x\rightarrow\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\rightarrow\infty}\frac{x^{7}}{e^{x^{6}}}=\lim_{x\rightarrow\infty}\frac{7x^{6}}{6x^{5}e^{x^{6}}}=\lim_{x\rightarrow\infty}\frac{7x}{6e^{x^{6}}} ).

Step3: Analyze the new limit

As ( x\rightarrow\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\rightarrow\infty}\frac{7x}{6e^{x^{6}}} ), and rewrite it as ( \frac{7}{6}\lim_{x\rightarrow\infty}\frac{x}{e^{x^{6}}} ). Let ( t=x^{6} ), then ( x = t^{\frac{1}{6}} ) and as ( x\rightarrow\infty ), ( t\rightarrow\infty ). So, ( \frac{7}{6}\lim_{x\rightarrow\infty}\frac{x}{e^{x^{6}}}=\frac{7}{6}\lim_{t\rightarrow\infty}\frac{t^{\frac{1}{6}}}{e^{t}} ). We know that for any positive real number ( a ), ( \lim_{t\rightarrow\infty}\frac{t^{a}}{e^{t}} = 0 ) (by repeatedly applying L'Hospital's Rule ( \lceil a\rceil+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:

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