Calculating the limit $\lim_{x \to \infty} \frac{(x+2)^b-(x+1)^b }{(x+1)^b-x^b}$ I want to prove/find the following limit:
$$\lim_{x \to \infty} \frac{(x+2)^b-(x+1)^b }{(x+1)^b-x^b}$$
where $1<b\leq1.5$ is fixed.
if trying different values for $b$, wolframalpha suggests that the limit is 1. But if so, I would really like how to prove it, especially because the term has this nice structure.
Please help!
 A: If you set $x=\frac 1t$ and consider $t\to0^+$ you get
$$\frac{(x+2)^b-(x+1)^b }{(x+1)^b-x^b} = \frac{(1+2t)^b-(1+t)^b }{(1+t)^b-1}$$
$$=\frac{(1+2t)^b-1 +1 - (1+t)^b }{t}\cdot \frac{1}{\frac{(1+t)^b-1}{t}}$$
Now, seeing that these are derivatives of $(1+2t)^b$ and $(1+t)^b$ at $t= 0$ you can proceed (taking into consideration that $b>1$)
$$\stackrel{t\to 0^+}{\longrightarrow}(2b-b)\cdot \frac 1{b}=1$$
A: $$
\frac{(x+2)^b-(x+1)^b}{(x+1)^b-x^b}\approx\frac{x^b+2bx^{b-1}-x^b-bx^{b-1}}{x^b+bx^{b-1}-x^b}=1.
$$
This is obtained using the binomial theorem, stopping at the first relevant terms.
A: $$\begin{align}
\lim_{x \to \infty} \frac{(x+2)^b-(x+1)^b }{(x+1)^b-x^b}
&\stackrel{(1)}= \lim_{x \to \infty} \frac{(1+\tfrac 2x)^b-(1+\tfrac 1x)^b }{(1+\tfrac 1x)^b-1}\\
&\stackrel{(2)}= \lim_{x \to 0^+} \frac{(1+2x)^b-(1+x)^b }{(1+x)^b-1}\\
&\stackrel{(3)}= \lim_{x \to 0^+} \frac{2b(1+2x)^{b-1}-b(1+x)^{b-1}}{b(1+x)^{b-1}}\\
&\stackrel{(4)}= \lim_{x \to 0^+} \frac{2(1+2x)^{b-1}-(1+x)^{b-1}}{(1+x)^{b-1}}\\
&\stackrel{(5)}= \frac{2-1}1 \stackrel{(6)}= 1
\end{align}$$
(1) is dividing by $x^b$
(2) is using $1/x\to 0^+$ instead of $x\to\infty$
(3) is l'Hospital which may be used because numerator and denominator approach 0 (and $b\neq 0$ hence derivative is not ln). 
(4) is shortening out $b\neq 0$
(5) this is carrying out the 3 lim $*^{b-1}$ individually and using that $(a,b,c)\mapsto (a-b)/c$ is continuous for $c\neq 0$
(6) this is left to the reader (I had this one wrongly 2 in the 1st version)
A: Set $y:=x+1$;
MVT:
Numerator: $f(y)=y^b.$
$f(y+1)-f(y)=f'(t)\cdot 1=bt^{b-1},$ $t \in (y,y+1)$.
Denominator:
$f(y)-f(y-1)=bs^{b-1}$, $s \in (y-1,y)$.
$\dfrac{y^{b-1}}{y^{b-1}} \lt \dfrac{f(y+1)-f(y)}{f(y)-f(y-1)} \lt$
$\dfrac{(y+1)^{b-1}}{(y-1)^{b-1}}=(\dfrac{y+1}{y-1})^{b-1}$.
Take the limit.
A: Some tests using bc seem to indicate that a good approximation for
$$f(x,b) = {{(x+2)^b-(x+1)^b}\over{(x+1)^b-x^b}}$$
is
$$g(x,b) = 1 + {{b-1}\over{x}}$$
Since the next term involves $1\over{x^2}$,  this seems to be close.  Note that the restriction on $b$ does not seem to be needed.
Now I'm not sure how to prove $$\lim_{x \to \infty}{f(x,b)} = \lim_{x\to\infty}g(x,b)$$ but $$\lim_{x\to\infty}g(x,b) \to 1$$ is easy.
