Let $\lambda\in(0,1)$, $h=\lfloor(i+1)\lambda\rfloor $. Show $$ \lim_{i\to\infty}\binom{i}{h}(1-\lambda)^{(i-h)}\lambda^{h}=0 $$
I was able to prove that for $\lambda=\frac{1}{2}$ using approximation for central binomial coefficent $\binom{2n}{n}\approx\frac{4^n}{\sqrt{\pi n}}$. But not the other cases.
Hint: First note that $h\to \infty$ and $i-h\to \infty$ as $i\to \infty$. Apply the Stirling approximation for the binomial $$ \binom{i}{h} =\frac{i!}{h! (i-h)!}\approx \sqrt \frac{i}{2 \pi h (h-i)} \left(\frac{i}{i-h}\right)^i \left(\frac{i-h}{h}\right)^h $$
And write $h=(i+1)\lambda-\epsilon$ with $0\le \epsilon <1$, replace and evaluate the limit.