Given Information:
We are given that the function f: R -> R is such that f" is continuous on R and f(0) = 1, f'(0) = 0, f"(0) = -1.

Required:
We need to find the limit of the expression lim(n->∞) (f(√2/x))^x as x tends to infinity.

Solution:

Step 1: Finding f'(x) and f''(x)
Since f'(0) = 0, we can use the Taylor series expansion to write:
f(x) = f(0) + f'(0)x + (1/2)f''(0)x^2 + ...

Given that f(0) = 1 and f'(0) = 0, the expansion becomes:
f(x) = 1 + (1/2)f''(0)x^2 + ...

Taking the derivative of f(x), we get:
f'(x) = 0 + f''(0)x + ...

Therefore, f'(x) = f''(0)x.

Step 2: Expressing f(√2/x) in terms of f(x)
Let's substitute x = √2/x in the Taylor series expansion of f(x):
f(√2/x) = 1 + (1/2)f''(0)(√2/x)^2 + ...

Simplifying the expression, we get:
f(√2/x) = 1 + (1/2)f''(0)(2/x) + ...

Since f''(0) = -1, the expression becomes:
f(√2/x) = 1 - (1/x) + ...

Step 3: Evaluating the limit
Now, let's find the limit of (f(√2/x))^x as x tends to infinity:
lim(x->∞) (f(√2/x))^x = lim(x->∞) (1 - (1/x) + ...)^x

Taking the natural logarithm of both sides, we have:
ln(lim(x->∞) (f(√2/x))^x) = ln(lim(x->∞) (1 - (1/x) + ...)^x)

Using the property ln(a^b) = b ln(a), the expression becomes:
ln(lim(x->∞) (f(√2/x))^x) = x ln(1 - (1/x) + ...)

As x tends to infinity, 1/x approaches 0, and ln(1 - (1/x) + ...) approaches ln(1) = 0.

Therefore, we have:
ln(lim(x->∞) (f(√2/x))^x) = x * 0 = 0

Taking the exponential of both sides, we get:
lim(x->∞) (f(√2/x))^x = e^0 = 1

Conclusion:
The limit of the expression lim(x->∞) (f(√2/x))^x is equal to 1.