Limit Question?

Question

Find the following:

lim x->0 of (1 - 2x) ^ (1/x)


(a) +infinity
(b) 1
(c) 1/2
(d) 2/e
(e) e^(-2)

Please explain your answer

I think I got it... someone check this please:

lim x->0 (1-2x) ^ (1/x)

let 1/y = -2x

as x->0, y-> - infinity

so lim y-> - infinity (1+ 1/y) ^ (-2y)

so 1/(lim y-> - infinity [1+1/y]^-y])^2

so 1/(e)^2

so e^(-2)

Answer 1

y = lim x→0 (1 - 2x)^(1/x)
ln y = lim x→0 ln {(1 - 2x)^(1/x)}
ln y = lim x→0 (1/x)ln (1 - 2x)
ln y = lim x→0 [ln (1 - 2x)]/x

Now we have an indeterminant form of 0/0 so we can apply L'Hospital's rule and take the derivative of the numerator and denominator and that will equal the same limit as the original.

ln y = lim x→0 [ln (1 - 2x)]/x
ln y = lim x→0 [-2/(1 - 2x)]/1
ln y = lim x→0 [-2/(1 - 2x)] = -2
y = e^(-2)

The answer is (e) e^(-2)

You are correct.

Answer 2

lim x->0 of (1 - 2x)^(1/x) =
lim x->0 of e^ln[(1 - 2x)^(1/x)] =
lim x->0 of e^(1/x)ln(1 - 2x) =
lim x->0 of e^[ln(1 - 2x)]/x =
lim x->0 of e^2(1/(1 - 2x)) =
e^2(1/(1 - 2)) =
e^2(1/-1) =
e^-2

Answer 3

lim x->0 (1-2x)^(1/x) = lim x->0 exp [ln (1-2x)^(1/x)]

= lim exp [ln (1-2x)]/x

Differentiate the top and bottom of the exponentiated term, then,

lim exp [-2/(1-2x)]
= exp (-2)

so the answer is e)

Answer 4

lim x->0 of (1 - 2x) ^ (1/x)

=lim x->0 of (1 - 2x) ^ (1/(-2x)^(-2)
=e^(-2)
Answer : (e)

Answer 5

1.

Answer 6

(1 - 2x)^(1/x) is the same as

e^(1/x)ln(1 - 2x)

e^(ln (1 - 2x) / x)

Therefore, taking the limit of this, we have

e^ (lim (ln (1 - 2x) / x) )

Using L'Hospital's rule,

e^ (lim (-2/(1 - 2x) ) = e^0 = 1

Answer 7

lim(1-2x)^(1/x) = (1-2x)^(2/2x) =((1-2x)^(-1/2x))^-2
The base tends to e so the limit is e^-2