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 . Just go ahead and divide it out: divided by 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 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 which does not exist.
1. Rewrite the term of the function:
2. Do long division:
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