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 $1&lt;b\leq1.5$ 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!
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 $b&gt;1$)

$$\stackrel{t\to 0^+}{\longrightarrow}(2b-b)\cdot \frac 1{b}=1$$
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.