What is the limit as x approaches infinity of (4x^2+5x-1)/(x^2-7x+29) ?

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limit as x approaches infinity of (4x^2+5x-1)/(x^2-7x+29)

x \to \infty lim (\frac{4 x ^{2} + 5 x - 1}{x ^{2} - 7 x + 29})

4

Solution steps

x \to \infty lim (\frac{4 x ^{2} + 5 x - 1}{x ^{2} - 7 x + 29})

Divide by highest denominator power:

\frac{4 + \frac{5}{x} - \frac{1}{x ^{2}}}{1 - \frac{7}{x} + \frac{29}{x ^{2}}}

= x \to \infty lim (\frac{4 + \frac{5}{x} - \frac{1}{x ^{2}}}{1 - \frac{7}{x} + \frac{29}{x ^{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{lim _{x \to \infty} ( 4 + \frac{5}{x} - \frac{1}{x ^{2}} )}{lim _{x \to \infty} ( 1 - \frac{7}{x} + \frac{29}{x ^{2}} )}

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

x \to \infty lim (1 - \frac{7}{x} + \frac{29}{x ^{2}}) = 1

= \frac{4}{1}

Simplify

= 4

x \to 1 lim (\frac{6 x + 1}{x - 1})

x \to 0 lim (\frac{tan ( x )}{x} cos (x))

x \to \infty lim ((\frac{1 + 9}{x})^{(\frac{x}{8})})

x \to \infty lim (- \infty + ln (2))

x \to 0 lim (\frac{x + π sec ( x )}{x ^{2} - π ^{2}})

What is the limit as x approaches infinity of (4x^2+5x-1)/(x^2-7x+29) ?
The limit as x approaches infinity of (4x^2+5x-1)/(x^2-7x+29) is 4