For fixed $x \geq 0$, find $\lim\limits_{n\to\infty}1-\left(\frac{n-\lambda}{n}\right)^{nx}$ 
For fixed $x \geq 0$, find $\lim\limits_{n\to\infty}1-\left(\frac{n-\lambda}{n}\right)^{nx}.$

Clearly, the object of interest is $\lim\limits_{n\to\infty}(\frac{n-\lambda}{n})^{nx}=\lim\limits_{n\to\infty}(1-\frac{\lambda}{n})^{nx}$. 
This closely resembles
$\lim\limits_{n\to\infty}(1-\frac{1}{n})^n=\frac{1}{e}$.
Otherwise, I am lost.
 A: Since $x\ge0$ is fixed, if
$$
\lim_{n\to\infty}\left(1-\frac{\lambda}{n}\right)^{\!n}=l
$$
exists, you also have
$$
\lim_{n\to\infty}\left(1-\frac{\lambda}{n}\right)^{\!nx}=
\lim_{n\to\infty}
  \left(
    \left(1-\frac{\lambda}{n}\right)^{\!n}
  \right)^{\!x}=
l^x
$$
because $t\mapsto t^x$ is continuous on $\mathbb{R}$ (if you don't trust in $0^0=1$, do a separate case for $x=0$, which however is trivial).
It's well known that $l$ exists and is finite: indeed $l=e^{-\lambda}$. Thus your limit is
$$
1-e^{-\lambda x}
$$
A: The approach is the standard one in the case of limits of the type $1^\infty$: $\Big( 1- \frac \lambda n \Big)^{nx} = \Big[ \Big( 1- \frac \lambda n \Big) ^{- \frac n \lambda} \Big] ^{-\frac \lambda n nx}$. The quantity between square brakets is guaranteed to tend to $\Bbb e$, so you only have to compute the limit of the exponent, which in this case is $-\lambda x$. Therefore, your limit is $\Bbb e ^{-\lambda x}$. This is for the case $\lambda \ne 0$, so that you may be able to write the fraction $\frac n \lambda$. If $\lambda = 0$, then the limit is trivially $1$.
