Solve ;limit as x approaches zero (4-2x^2)1/x

Answer 1

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.