What is the limit as x approaches-infinity of (8x^2+7x^{1/3})/(sqrt(16x^4+6)) ?

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limit as x approaches-infinity of (8x^2+7x^{1/3})/(sqrt(16x^4+6))

x \to - \infty lim (\frac{8 x ^{2} + 7 x ^{\frac{1}{3}}}{16 x ^{4} + 6})

2

Solution steps

x \to - \infty lim (\frac{8 x ^{2} + 7 x ^{\frac{1}{3}}}{16 x ^{4} + 6})

Apply L'Hopital's Rule

: x \to - \infty lim \frac{\frac{7 + 48 x ^{\frac{5}{3}}}{3 x ^{\frac{2}{3}}}}{\frac{32 x ^{3}}{( 16 x ^{4} + 6 ) ^{\frac{1}{2}}}}

= x \to - \infty lim \frac{\frac{7 + 48 x ^{\frac{5}{3}}}{3 x ^{\frac{2}{3}}}}{\frac{32 x ^{3}}{( 16 x ^{4} + 6 ) ^{\frac{1}{2}}}}

Simplify

\frac{\frac{7 + 48 x ^{\frac{5}{3}}}{3 x ^{\frac{2}{3}}}}{\frac{32 x ^{3}}{( 16 x ^{4} + 6 ) ^{\frac{1}{2}}}} : \frac{( 7 + 48 x ^{\frac{5}{3}} ) ( 16 x ^{4} + 6 ) ^{\frac{1}{2}}}{96 x ^{\frac{11}{3}}}

= x \to - \infty lim \frac{( 7 + 48 x ^{\frac{5}{3}} ) ( 16 x ^{4} + 6 ) ^{\frac{1}{2}}}{96 x ^{\frac{11}{3}}}

Apply L'Hopital's Rule

: x \to - \infty lim \frac{\frac{224 x ^{3} + 2816 x ^{\frac{14}{3}} + 480 x ^{\frac{2}{3}}}{( 16 x ^{4} + 6 ) ^{\frac{1}{2}}}}{352 x ^{\frac{8}{3}}}

= x \to - \infty lim \frac{\frac{224 x ^{3} + 2816 x ^{\frac{14}{3}} + 480 x ^{\frac{2}{3}}}{( 16 x ^{4} + 6 ) ^{\frac{1}{2}}}}{352 x ^{\frac{8}{3}}}

Simplify

\frac{\frac{224 x ^{3} + 2816 x ^{\frac{14}{3}} + 480 x ^{\frac{2}{3}}}{( 16 x ^{4} + 6 ) ^{\frac{1}{2}}}}{352 x ^{\frac{8}{3}}} : \frac{7 x ^{3} + 88 x ^{\frac{14}{3}} + 15 x ^{\frac{2}{3}}}{11 x ^{\frac{8}{3}} ( 16 x ^{4} + 6 ) ^{\frac{1}{2}}}

= x \to - \infty lim (\frac{7 x ^{3} + 88 x ^{\frac{14}{3}} + 15 x ^{\frac{2}{3}}}{11 x ^{\frac{8}{3}} ( 16 x ^{4} + 6 ) ^{\frac{1}{2}}})

x \to a lim [c \cdot f (x)] = c \cdot x \to a lim f (x)

= \frac{1}{11} \cdot x \to - \infty lim (\frac{7 x ^{3} + 88 x ^{\frac{14}{3}} + 15 x ^{\frac{2}{3}}}{x ^{\frac{8}{3}} ( 16 x ^{4} + 6 ) ^{\frac{1}{2}}})

Divide by

x^{\frac{14}{3}} : \frac{\frac{7}{x ^{\frac{5}{3}}} + 88 + \frac{15}{x ^{4}}}{( 16 + \frac{6}{x ^{4}} ) ^{\frac{1}{2}}}

= \frac{1}{11} \cdot x \to - \infty lim \frac{\frac{7}{x ^{\frac{5}{3}}} + 88 + \frac{15}{x ^{4}}}{( 16 + \frac{6}{x ^{4}} ) ^{\frac{1}{2}}}

x \to a lim [\frac{f ( x )}{g ( x )}] = \frac{lim _{x \to a} f ( x )}{lim _{x \to a} g ( x )}, x \to a lim g (x) \neq = 0

With the exception of indeterminate form

= \frac{1}{11} \cdot \frac{lim _{x \to - \infty} ( \frac{7}{x ^{\frac{5}{3}}} + 88 + \frac{15}{x ^{4}} )}{lim _{x \to - \infty} ( ( 16 + \frac{6}{x ^{4}} ) ^{\frac{1}{2}} )}

x \to - \infty lim (\frac{7}{x ^{\frac{5}{3}}} + 88 + \frac{15}{x ^{4}}) = 88

x \to - \infty lim ((16 + \frac{6}{x ^{4}})^{\frac{1}{2}}) = 4

= \frac{1}{11} \cdot \frac{88}{4}

Simplify

\frac{1}{11} \cdot \frac{88}{4} : 2

= 2

(cos (t))^{^{'}}

\frac{d}{d x} ((x e^{2 x} + x^{3})^{6})

\int \frac{1}{2 x - x ^{2}} d x

\int_{- 1}^{0} (x^{3} - x^{2} - 2 x) d x

What is the limit as x approaches-infinity of (8x^2+7x^{1/3})/(sqrt(16x^4+6)) ?
The limit as x approaches-infinity of (8x^2+7x^{1/3})/(sqrt(16x^4+6)) is 2