What is the limit as x approaches-infinity of 3x+sqrt(9x^2+7x-8) ?

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limit as x approaches-infinity of 3x+sqrt(9x^2+7x-8)

x \to - \infty lim (3 x + 9 x^{2} + 7 x - 8)

- \frac{7}{6}

Decimal

- 1.16666 \dots

Solution steps

x \to - \infty lim (3 x + 9 x^{2} + 7 x - 8)

Multiply by the conjugate of

3 x + 9 x^{2} + 7 x - 8 : \frac{- 7 x + 8}{3 x - 9 x ^{2} + 7 x - 8}

= x \to - \infty lim (\frac{- 7 x + 8}{3 x - 9 x ^{2} + 7 x - 8})

Divide by highest denominator power:

\frac{- 7 + \frac{8}{x}}{3 + 9 + \frac{7}{x} - \frac{8}{x ^{2}}}

= x \to - \infty lim \frac{- 7 + \frac{8}{x}}{3 + 9 + \frac{7}{x} - \frac{8}{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} ( - 7 + \frac{8}{x} )}{lim _{x \to - \infty} ( 3 + 9 + \frac{7}{x} - \frac{8}{x ^{2}} )}

x \to - \infty lim (- 7 + \frac{8}{x}) = - 7

x \to - \infty lim (3 + 9 + \frac{7}{x} - \frac{8}{x ^{2}}) = 6

= \frac{- 7}{6}

Simplify

= - \frac{7}{6}

\int 16 x d x

\int \frac{40 x}{5 x ^{2} + e} d x

x \to 0 + lim (tan (x) ln (x))

\frac{d}{d x} (3 x + x)

y^{^{''}} + 4 y^{^{'}} + 5 y = 5 x + 3 e - x

What is the limit as x approaches-infinity of 3x+sqrt(9x^2+7x-8) ?
The limit as x approaches-infinity of 3x+sqrt(9x^2+7x-8) is -7/6