How to say limit of this expression is finite

Using Taylor Theorem, it is easy to show that $log(1-x)=-x+O(x^2)$, from which we have

$$begin{align} left(1-frac{t^2}{2r}+Oleft(r^{-3/2}right)right)^{-r}&=e^{-rlogleft(1-frac{t^2}{2r}+Oleft(r^{-3/2}right)right)}\\ &=e^{t^2/2}e^{-O(r^{-1/2})} end{align}$$

Therefore, we see that

$$lim_{rto infty }left(1-frac{t^2}{2r}+Oleft(r^{-3/2}right)right)^{-r}=e^{t^2/2}$$

as was to be shown!

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EDIT:

The OP was interested in proceeding with a binomial expansion. To that end, we now use a generalized binomial expansion of the term if interest to write

$$begin{align} left(1-frac{t^2}{2r}+Oleft(r^{-3/2}right)right)^{-r}&=sum_{k=0}^infty binom{-r}{k}left(1-frac{t^2}{2r}right)^{-r-k}Oleft(r^{-3/2}right)^ktag1 end{align}$$

where the generalized binomial coefficient, $binom{-r}{k}$, is given by

$$binom{-r}{k}=frac{-r(-r-1)cdots (-r-k+1)}{k!}$$

Owing to the uniform convergence of the series on the right-hand side of $(1)$, we can interchange the order of the limit and the summation to obtain

$$begin{align} lim_{rtoinfty}sum_{k=0}^infty binom{-r}{k}left(1-frac{t^2}{2r}right)^{-r-k}Oleft(r^{-3/2}right)^k&=sum_{k=0}^infty lim_{rto infty}left(binom{-r}{k}left(1-frac{t^2}{2r}right)^{-r-k}Oleft(r^{-3/2}right)^kright)\\ &=e^{t^2/2} end{align}$$

since for $kne 0$,

$$binom{-r}{k}Oleft(r^{-3/2}right)^k=Oleft(r^{-1/2}right)^kto 0$$

as $rto infty$ and

$$lim_{rto infty}left(1-frac{t^2}{2r}right)^{-r-k}=e^{t^2/2}$$