To solve this limit, we can use logarithmic differentiation.
First, we rewrite the expression as:
(4-2x^2)^(1/x) = e^(ln(4-2x^2) / x)
Then, using properties of logarithms, we can rewrite the expression inside the natural logarithm:
ln(4-2x^2) = ln(2(2-x)(2+x))
= ln(2) + ln(2-x) + ln(2+x)
Now, we can take the natural logarithm of the entire original expression:
ln((4-2x^2)^(1/x)) = ln(e^(ln(4-2x^2) / x))
= ln(4-2x^2) / x
= (ln(2) + ln(2-x) + ln(2+x)) / x
As x approaches zero, the denominator of this expression approaches zero, so we have an indeterminate form of infinity over infinity. We can use L'Hopital's rule to evaluate this limit:
lim x->0 (ln(2) + ln(2-x) + ln(2+x)) / x
= lim x->0 ((1/(2-x)) + (-1/(2+x)))
= 1/2 - 1/2
= 0
Therefore, the limit as x approaches zero of (4-2x^2)^(1/x) is e^0, which simplifies to 1.
Answer: 1.
Solve ;limit as x approaches zero (4-2x^2)1/x
1 answer