$$\lim \limits_{n \to \infty}(n^5+4n^3)^{1/5}-n=?$$
I see that $$\lim_{n \to \infty}(n^5+4n^3)^{1/5}-n=\lim_{n \to \infty}n[(1+ \frac {4}{n^2})^{1/5}-1]=\lim_{z \to 0} \frac {1}{z}[(1+ {4}{z^2})^{1/5}-1]$$,where $n=\frac {1}{z}$. Now I do not know how to proceed.
Can someone point me in the right direction? Thanks in advance for your time.
Two methods:
Here, the identity $(a-b)(a^4+a^3b+a^2b^2+ab^3+b^4)=a^5-b^5$ with $a=(n^5+4n^3)^{1/5}$ and $b=n$ yields $a^5-b^5=4n^3$. Furthermore, $a\sim b\sim n$ hence each term in the sum $a^4+a^3b+a^2b^2+ab^3+b^4$ is equivalent to $n^4$. Can you finish?
Here, $(1+x)^{1/5}=1+\frac15x+o(x)$ when $x\to0$ hence $(1+4/n^2)^{1/5}=1+O(1/n^2)$. Furthermore, $(n^5+4n^3)^{1/5}=n(1+4/n^2)^{1/5}$. Can you finish?
Let us stare at the expression $$\lim_{z \to 0} \frac {1}{z}[(1+ {4}{z^2})^{1/5}-1]$$ that you reached, or, to use letters that are more familiar from first-year calculus, let us stare at $$\lim_{h \to 0} \frac{(1+ {4}{h^2})^{1/5}-1}{h}.$$ This is the standard expression for the derivative of $f(x)=(1+4x^2)^{1/5}$ at $x=0$.
Now calculate the derivative in the usual way.
Hint:
$$ (x^{1/5}-1)(1+x^{1/5}+x^{2/5}+x^{3/5}+x^{4/5}) = x-1 $$ Rearrange and substitute for $x=1+4/n^2$.
Hint: $\frac{(1+4z^2)^{1/5}-1}{z}$ is a $\left[\frac{0}{0}\right]$ expression as $z\to 0$; it suffices to use de l'Hospital once, and you should obtain your answer.
An answer without L'Hospital's: the binomial theorem states that $\left(1+z\right)^t = 1+tz+\dfrac{t(t-1)}{2}z^2+\ldots$. This expansion holds perfectly well even if $t$ isn't a whole number; the expression just becomes an infinite rather than a finite sum. Plug in the appropriate values for $z$ and $t$ here and subtract off the constant term, and your limit should fall out fairly easily.
