Question

find \\( \lim _{x \
ightarrow \infty} h(x) \\), if it exists. (if an answer does not exist, enter dne.)

\\( f(x)=7 x^{3}-6 \\)

(a) \\( h(x)=\frac{f(x)}{x^{2}} \\)

\\( \lim _{x \
ightarrow \infty} h(x)= \\)

(b) \\( h(x)=\frac{f(x)}{x^{3}} \\)

\\( \lim _{x \
ightarrow \infty} h(x)= \\)

(c) \\( h(x)=\frac{f(x)}{x^{4}} \\)

\\( \lim _{x \
ightarrow \infty} h(x)= \\)

Explanation:

Step1: Simplify \( h(x) \) for part (a)

Given \( h(x)=\frac{7x^{3}-6}{x^{2}} = 7x-\frac{6}{x^{2}} \).
As \( x
ightarrow\infty \), \( \lim_{x
ightarrow\infty}(7x-\frac{6}{x^{2}})=\lim_{x
ightarrow\infty}(7x)-\lim_{x
ightarrow\infty}(\frac{6}{x^{2}}) \).
Since \( \lim_{x
ightarrow\infty}\frac{6}{x^{2}} = 0 \) and \( \lim_{x
ightarrow\infty}(7x)=\infty \).

Step2: Simplify \( h(x) \) for part (b)

Given \( h(x)=\frac{7x^{3}-6}{x^{3}}=7-\frac{6}{x^{3}} \).
As \( x
ightarrow\infty \), \( \lim_{x
ightarrow\infty}(7 - \frac{6}{x^{3}})=\lim_{x
ightarrow\infty}(7)-\lim_{x
ightarrow\infty}(\frac{6}{x^{3}}) \).
Since \( \lim_{x
ightarrow\infty}\frac{6}{x^{3}}=0 \), then \( \lim_{x
ightarrow\infty}(7-\frac{6}{x^{3}})=7 \).

Step3: Simplify \( h(x) \) for part (c)

Given \( h(x)=\frac{7x^{3}-6}{x^{4}}=\frac{7}{x}-\frac{6}{x^{4}} \).
As \( x
ightarrow\infty \), \( \lim_{x
ightarrow\infty}(\frac{7}{x}-\frac{6}{x^{4}})=\lim_{x
ightarrow\infty}(\frac{7}{x})-\lim_{x
ightarrow\infty}(\frac{6}{x^{4}}) \).
Since \( \lim_{x
ightarrow\infty}\frac{7}{x}=0 \) and \( \lim_{x
ightarrow\infty}\frac{6}{x^{4}} = 0 \).

Answer:

(a) \( \infty \)
(b) \( 7 \)
(c) \( 0 \)