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Please how can I calculate limit of $n^2 (sqrt[n]{n+1}-sqrt[n]{n})$ without using Taylor's expansion.

If $a>b>0$ we have
$$ a^n-b^n = (a-b)(a^{n-1}+a^{n-2}b+ldots+a b^{n-2} + b^{n-1})in(n(a-b)b^{n-1},n(a-b)a^{n-1}).$$
By considering $a=sqrt[n]{n+1}$ and $b=sqrt[n]{n}$ it follows that
$$ frac{sqrt[n]{n+1}}{n(n+1)}leq sqrt[n]{n+1}-sqrt[n]{n} leq frac{sqrt[n]{n}}{n^2}$$
and since both $sqrt[n]{n}$ and $sqrt[n]{n+1}$ converge to $1$ as $nto +infty$, the wanted limit is $color{red}{1}$ by squeezing.

By the MVT we have

$$(n+1)^{1/n}-n^{1/n} = (1/n)cdot c_n^{1/n-1}cdot 1$$

for some $c_nin(n,n+1).$ The right side is less than

$$(1/n)n^{1/n-1}= (1/n^2)n^{1/n}.$$

If we multiply this by $n^2$ we get an expression that $to 1.$ The argument from below is similar. The limit of our expression is thus $1$ by the squeeze principle.

Since
$a^n-b^n
=(a-b)sum_{k=0}^{n-1} a^kb^{n-1-k}
$,
$a-b
=(a^{1/n}-b^{1/n})sum_{k=0}^{n-1} a^{k/n}b^{(n-1-k)/n}
$.

Letting
$a=n+1, b=n$,
this becomes
$1
=((n+1)^{1/n}-n^{1/n})sum_{k=0}^{n-1} (n+1)^{k/n}n^{(n-1-k)/n}
$.

$begin{array}\
s(n)
&=sum_{k=0}^{n-1} (n+1)^{k/n}n^{(n-1-k)/n}\
&=sum_{k=0}^{n-1} (1+1/n)^{k/n}n^{k/n}n^{(n-1-k)/n}\
&=sum_{k=0}^{n-1} (1+1/n)^{k/n}n^{1-1/n}\
&=n^{1-1/n}sum_{k=0}^{n-1} (1+1/n)^{k/n}\
text{so}\
s(n)
&gt n^{1-1/n}n\
&=n^2n^{-1/n}\
text{and}\
s(n)
&lt ncdot n(1+1/n)\
&=n^2+n\
end{array}
$

Therefore
$(n+1)^{1/n}-n^{1/n}
=dfrac1{s(n)}
lt dfrac{n^{1/n}}{n^2}
$
and
$(n+1)^{1/n}-n^{1/n}
=dfrac1{s(n)}
gt dfrac1{n^2+n}
$
so that
$dfrac1{1+1/n}
lt n^2((n+1)^{1/n}-n^{1/n})
lt n^{1/n}
lt 1+dfrac{3}{sqrt{n}}
$.

Note:
To show $n^{1/n} to 1$,
by Bernoulli's inequality,
$(1+dfrac1{sqrt{n}})^n
ge 1+dfrac{n}{sqrt{n}}
gt sqrt{n}
=n^{1/2}
$.

Raising to the
$2/n$ power,
$(1+dfrac1{sqrt{n}})^2
gt n^{1/n}
$
so
$n^{1/n}
lt (1+dfrac1{sqrt{n}})^2
= 1+dfrac{2}{sqrt{n}}+dfrac1{n}
lt 1+dfrac{3}{sqrt{n}}
to 1$.

Just take $n^{1/n}$ as a common factor from both terms and note that $n^{1/n}$ tends to $1$ so that the desired limit is equal to the limit of $$n^2{(1+n^{-1})^{1/n} - 1} $$ Since $a_n=(1+n^{-1})^{1/n}to 1$ we can write the above expression as $$n^2log a_ncdotfrac{a_n-1}{log a_n} $$ and the last factor tends to $1$ so that the desired limit is equal to the limit of $$nlog(1+n^{-1})$$ which is $1$.

We have that by $x=frac1n to 0^+$

$$n^2(sqrt[n]{n+1}-sqrt[n]{n})=frac{left(1+frac1xright)^x-left(frac1xright)^x}{x^2}=left(frac1xright)^xfrac{left(1+xright)^x-1}{x^2}$$

and since $left(frac1xright)^xto 1$ by l'Hopital we obtain

$$lim_{xto 0}frac{left(1+xright)^x-1}{x^2}=lim_{xto 0}frac{left(1+xright)^xleft(log(1+x)+frac x{1+x}right)}{2x}=$$

$$=lim_{xto 0}frac{left(1+xright)^xleft(log(1+x)+frac x{1+x}right)^2+(1+x)^xleft(frac1{1+x}+frac1{(1+x)^2}right)}{2}=1$$

We have that

$$sqrt[n]{n+1}=sqrt[n]{n}cdot left(1+frac1nright)^frac1n$$

and by $(1)$ and Bernoully inequality

$$left(1+frac1{n^2}-frac1{2n^3}right)le left(1+frac1nright)^frac1nle left(1+frac1{n^2}right)$$

and then

$$sqrt[n]{n}+frac{sqrt[n]{n}}{n^2}-frac{sqrt[n]{n}}{2n^3}lesqrt[n]{n+1}lesqrt[n]{n}+frac{sqrt[n]{n}}{n^2}$$

$$frac{sqrt[n]{n}}{n^2}-frac{sqrt[n]{n}}{2n^3}lesqrt[n]{n+1}-sqrt[n]{n}lefrac{sqrt[n]{n}}{n^2}$$

$$sqrt[n]{n}-frac{sqrt[n]{n}}{2n}le n^2(sqrt[n]{n+1}-sqrt[n]{n})lesqrt[n]{n}$$

then conclude by squeeze theorem.

Proof of (1)

By $x=frac1n to 0^+$

$$left(1+frac1{n^2}-frac1{2n^3}right)le left(1+frac1nright)^frac1n iff 1+x^2-frac12 x^3 le (1+x)^x$$

follows from

$$(1+x)^x=e^{xlog(1+x)} ge e^{x(x-frac12x^2)}=e^{x^2-frac12x^3}ge 1+x^2-frac12 x^3$$