What is the limit as x approaches-infinity of ((sqrt(x^4-1)))/((x^3-1)) ?

The limit as x approaches-infinity of ((sqrt(x^4-1)))/((x^3-1)) is 0

Popular >

Evaluate Using the Squeeze Theorem limit as x approaches negative infinity of ( square root of x^4-1)/(x^3-1)

x \to - \infty lim (\frac{( x ^{4} - 1 )}{( x ^{3} - 1 )})

0

Solution steps

x \to - \infty lim (\frac{x ^{4} - 1}{x ^{3} - 1})

Divide by highest denominator power:

\frac{\frac{x ^{4} - 1}{x x ^{4}}}{1 - \frac{1}{x ^{3}}}

= x \to - \infty lim \frac{\frac{x ^{4} - 1}{x x ^{4}}}{1 - \frac{1}{x ^{3}}}

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{x ^{4} - 1}{x x ^{4}} )}{lim _{x \to - \infty} ( 1 - \frac{1}{x ^{3}} )}

x \to - \infty lim (\frac{x ^{4} - 1}{x x ^{4}}) = 0

x \to - \infty lim (1 - \frac{1}{x ^{3}}) = 1

= \frac{0}{1}

Apply rule

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

= 0

x \to 1 - lim (\frac{( x - 1 )}{( x ^{2} - 4 )})

(- 3, 3), (5, 9)

\int \frac{( ln ( x ) )}{x} d x

\int (\frac{( x ^{2} - 9 )}{( x ^{3} )}) d x

\int (4 - x^{2}) d x

What is the limit as x approaches-infinity of ((sqrt(x^4-1)))/((x^3-1)) ?
The limit as x approaches-infinity of ((sqrt(x^4-1)))/((x^3-1)) is 0