What is the limit as x approaches infinity of (x^2-1)/(8x^2-5) ?

The limit as x approaches infinity of (x^2-1)/(8x^2-5) is 1/8

limit as x approaches infinity of (x^2-1)/(8x^2-5)

x \to \infty lim (\frac{x ^{2} - 1}{8 x ^{2} - 5})

\frac{1}{8}

Decimal

0.125

Solution steps

x \to \infty lim (\frac{x ^{2} - 1}{8 x ^{2} - 5})

Divide by highest denominator power:

\frac{1 - \frac{1}{x ^{2}}}{8 - \frac{5}{x ^{2}}}

= x \to \infty lim (\frac{1 - \frac{1}{x ^{2}}}{8 - \frac{5}{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} ( 1 - \frac{1}{x ^{2}} )}{lim _{x \to \infty} ( 8 - \frac{5}{x ^{2}} )}

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

x \to \infty lim (8 - \frac{5}{x ^{2}}) = 8

= \frac{1}{8}

x \to 0 lim (\frac{tan ( 4 x )}{sin ( 6 x )})

x \to - 2 lim (6)

x \to 0 lim (\frac{e ^{x} - 2}{sin ( 7 x )})

x \to 0 lim (tan^{x^{2}} (x))

x \to 5 lim (\frac{1}{2} (x + 3))

What is the limit as x approaches infinity of (x^2-1)/(8x^2-5) ?
The limit as x approaches infinity of (x^2-1)/(8x^2-5) is 1/8