$$\lim_{x \to -∞} \frac{\left(x^n+1\right)^{15}-\left(x^n-1\right)^{15}}{\left(x^n-3\right)^7+\left(x^n+2\right)^7} \ = -∞ $$ -
( We need to find the values of $n$ so this limit would equal $-∞$ ) I used some simplifications using the $a^n-b^n \\$ formulas and got $\lim_{x \to -∞} x^{2 n} \\$ )
Which can't be possible since $2n$ is an even number. i know for a fact that i can't put $n = 1/2$ because we haven't studied $a^k$ with $k$ part of the Q groups.
Any little hints would be so appreciated (just hints if i gave up i will make sure to tell you)
Hint:
Use asymptotic equivalents:
By the binomial formula, \begin{align} (x^n+1)^{15}-(x^n-1)^{15}&=x^{15n}+15x^{14 n}+ o\bigl(x^{14n}\bigr)-\Bigl(x^{15n}-15x^{14n}+ o\bigl(x^{14n}\bigr)\Bigr)\\ &= 30x^{14n}++ o\bigl(x^{14n}\bigr), \end{align} so that $\;(x^n+1)^{15}-(x^n-1)^{15}\sim_\infty 30x^{14n}$.
Similarly, $\;(x^n-3)^7+(x^n+2)^7\sim_\infty 2x^{7n}$,. Therefore $$\frac{(x^n+1)^{15}-(x^n-1)^{15}}{(x^n-3)^7+(x^n+2t)^7}\sim_\infty \frac{30x^{14n}}{2x^{7n}}=15x^{7n}.$$