I need help with this limits pls

[uncut inlet]

It is also stated that I ought to find the different values of it depending on a and b

[alpine tusk]

What is n approaching?

[uncut inlet]

Plus infinite

[alpine tusk]

There 3 cases

1. n² remains based on a
2. n² vanishes based on a
3. n² and n vanish based in a and b

[uncut inlet]

Thanks

[tulip kestrel]

$\lim_{n\to\infty} \frac{(2a - 8)n^2 + bn - 6}{-2n + 5}$

$= \lim_{n\to\infty} \frac{\cfrac{(2a - 8)n^2}{n} + \cfrac{bn}{n} - \cfrac{6}{n}}{\cfrac{-2n}{n} + \cfrac{5}{n}}$

$= \lim_{n\to\infty} \frac{2(a - 4)n + b - \cfrac{6}{n}}{\cfrac{5}{n} - 2}$

If $a = 4$, and $b \neq 0$, the limit is $\lim_{n\to\infty} \frac{0 + b - 0}{0 - 2} = -\frac{b}{2}$

If $a = 4$, and $b = 0$, the limit is $\lim_{n\to\infty} \frac{0 + 0 - 0}{0 - 2} = 0$

If $a > 4$, the limit is $\lim_{n\to\infty} \frac{(+)n + b - 0}{0 - 2}$ which diverges to $-\infty$

If $a < 4$, the limit is $\lim_{n\to\infty} \frac{(-)n + b - 0}{0 - 2}$ which diverges to $+\infty$

[languid bridge]

Jay

[tulip kestrel]

@uncut inlet to let you know there is a more detailed response posted 🙂

[uncut inlet]

thanks man,I appreciate that

.close