Vrftsjtryk

Can someone help me calculating this limit?

$$lim_{x to infty} frac{ (1-frac{15}{x})^{15} -1 }{(1-frac{15}{x}) -1}$$

I can't use L'Hospital's rule.

As we knew from before $$a^n-1=(a-1)(a^{n-1}+ldots +1)$$therefore by substituting $u=1-{15over x}$ and $lim_{xto infty} u=1$ we obtain $$lim_{x to infty} frac{ (1-frac{15}{x})^{15} -1 }{(1-frac{15}{x}) -1}=lim _{uto 1}{u^{15}-1over u-1}=lim _{uto 1} u^{14}+ldots +1 =15$$

HINT

Recall that

$$a^{15}-1=(a-1)(a^{14}+a^{13}+ldots+a+1)$$

A totally mechanical way that doesn't involve recognising the cyclotomic polynomial:

Get rid of the ridiculous terms by making the substitution $u = 1-frac{15}{x}$, just as in Mostafa Ayaz's answer.

Then since limits to $1$ are less natural than limits to $0$, substitute again:

$$lim_{u to 1} frac{u^{15}-1}{u-1} = lim_{h to 0} frac{(h+1)^{15}-1}{h} = f'(0)$$
where $f(x) = (x+1)^{15}$.

Now simply differentiate using the chain rule.

$y=15/x$, $xnot =0$, and consider

$y rightarrow 0^+$.

Numerator (Binomial expansion):

$(1-y)^{15}-1=$

$small{1-15y +(15)(14)y^2/2! -..+..-y^{15} -1}.$

Denominator: $-y$

The limit $y rightarrow 0^+$ is?

Or you can also use the binomial theorem:
$$frac{left(1-frac{15}{x}right)^{15}-1}{left(1-frac{15}{x}right)-1}=frac{1-frac{15}{x}binom{15}{1}+rleft(frac{15}{x}right)-1}{-frac{15}{x}}=frac{frac{225}{x}+rleft(frac{15}{x}right)}{frac{15}{x}}=15+frac{rleft(frac{15}{x}right)}{frac{15}{x}}to15$$
Where $r$ is a polinomial with lowest degree of $2$. You can calculate it, but we don't need to do so. It's enough to know that it looks something like $rleft(frac{15}{x}right)=Afrac{1}{x^2}+Bfrac{1}{x^3}ldots$, so $frac{rleft(frac{15}{x}right)}{frac{15}{x}}to0$.