What is the limit as x approaches infinity of ((5^x+x^4))/((7^x+x^2)) ?

The limit as x approaches infinity of ((5^x+x^4))/((7^x+x^2)) is 0

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limit as x approaches infinity of ((5^x+x^4))/((7^x+x^2))

x \to \infty lim (\frac{( 5 ^{x} + x ^{4} )}{( 7 ^{x} + x ^{2} )})

0

Solution steps

x \to \infty lim (\frac{5 ^{x} + x ^{4}}{7 ^{x} + x ^{2}})

Divide by

7^{x} : \frac{( \frac{5}{7} ) ^{x} + \frac{x ^{4}}{7 ^{x}}}{1 + \frac{x ^{2}}{7 ^{x}}}

= x \to \infty lim (\frac{( \frac{5}{7} ) ^{x} + \frac{x ^{4}}{7 ^{x}}}{1 + \frac{x ^{2}}{7 ^{x}}})

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{lim _{x \to \infty} ( ( \frac{5}{7} ) ^{x} + \frac{x ^{4}}{7 ^{x}} )}{lim _{x \to \infty} ( 1 + \frac{x ^{2}}{7 ^{x}} )}

x \to \infty lim ((\frac{5}{7})^{x} + \frac{x ^{4}}{7 ^{x}}) = 0

x \to \infty lim (1 + \frac{x ^{2}}{7 ^{x}}) = 1

= \frac{0}{1}

Apply rule

\frac{0}{a} = 0, a \neq = 0

= 0

x \to - 3 - lim (\frac{x + 1}{x ^{2} - 9})

x \to π lim (sin^{2} (x))

x \to 0 lim (x^{2} + 3 x - 4)

x \to 4 lim ((4 - x) \cdot ln (x - 4))

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

What is the limit as x approaches infinity of ((5^x+x^4))/((7^x+x^2)) ?
The limit as x approaches infinity of ((5^x+x^4))/((7^x+x^2)) is 0