1. ## Slant asymptote

x^n - n^2
x^n-1 - n

Does this statement have a slant asymptote for all values of n, if n is a natural number

Everything in bold is in superscript

2. Originally Posted by shane99
x^n - n^2
x^n-1 - n

Does this statement have a slant asymptote for all values of n, if n is a natural number

Everything in bold is in superscript
I presum you mean $\frac{x^n- n^2}{x^{n-1}- n}$. Just go ahead and divide it out: $x^{n}- n^2$ divided by $x^{n-1}-n$ is equal to x+ a fraction with lower degree in the numerator than in the denominator. Thus, y= x+ n is a slant asymptote.

The only "technical" points are for n= 2, where the fraction is $\frac{x^2- 4}{x- 2}$ which is equal to x+ 2 for all x not equal to 2- would you call that an asymptote?- and n= 1 where the fraction is $\frac{x- 1}{1- 1}$ which does not exist.

3. Originally Posted by shane99
x^n - n^2
x^n-1 - n

Does this statement have a slant asymptote for all values of n, if n is a natural number

Everything in bold is in superscript
1. Rewrite the term of the function:

$f_n(x)=\dfrac{x^n-n^2}{x^{n-1}-n}=\dfrac{x^n-n^2}{\frac1x \cdot x^n-n} = \dfrac{x^{n+1}-n^2 x}{x^n - nx}$

2. Do long division:

$f_n(x)= \dfrac{x^{n+1}-n^2 x}{x^n - nx} = x-\dfrac{n^2 x}{x^n-nx} + \dfrac{nx^2}{x^n-nx}$

3. f_1(x) doesn't exist for n = 1 because the denominator would be zero.

f_2 (x) (that means n = 2) has the asymptote y = x+2

For all n > 2 the fractions in the term become zero if x approaches infinity. Therefore the asymptote is y = x.

EDIT: Beaten! Again