A tricky limit (indeterminate form) While tutoring I came upon this limit. I know that this limit is obviously 1, but how would I show this formally 
$$\lim_{\eta\rightarrow\infty}[(2\eta + 5)^x-(2\eta)^x + 1]$$ where $x\in (0,1)$
I've tried logarithms etc.  
 A: Hint:
$$
\lim_{\eta \to \infty}(2 \eta + 5)^x - (2 \eta)^x = 
\lim_{\eta \to \infty}
5\eta^{x-1}\frac{(2 + 5/\eta)^x - 2^x}{5/\eta}
$$

Alternative approach: make the substitution $t = 1/\eta$ to rewrite this as
$$
\lim_{t \to 0^+} (2/t + 5)^x - (2 /t)^x = 
\lim_{t \to 0^+} \frac{(2 + 5t)^x - 2^x}{t^x}
$$
Now, apply L'Hopital.
A: Another way than the one suggested by Omnomnomnom is the following. Forget about the $+1$ for a moment and consider
$$\begin{aligned}
(2\eta + 5)^x - (2\eta )^x & = (2\eta)^x\left(1+\frac{5}{2\eta}\right)^x-(2\eta)^x\\
& = (2\eta)^x\left(\left(1+\frac{5}{2\eta}\right)^x-1\right).
\end{aligned}$$
Then you can use that
$$\begin{aligned}
\lim_{n\to\infty}(2\eta)^x\left(\left(1+\frac{5}{2\eta}\right)^x-1\right)& = \lim_{n\to\infty}\frac{\left(1+\frac{5}{2\eta}\right)^x-1}{\frac{1}{(2\eta)^x}}
\end{aligned}
$$
And then you can infer Hospitals rule.
A: By the MVT
$$0<(2\eta+5)^x-(2\eta)^x=5\> x(2\eta +\tau)^{x-1}$$
for some $\tau\in\>]0,5[\>$. Since the RHS converges to $0$ when $\eta\to\infty$ the limit in question is $1$.
