Hi,

I am trying to figure out what the limit as h approaches 0 of (1-2h)^(1/h) is. I am unfamiliar with the process I am supposed to use to solve this limit. I have just been reasoning out this limit, but I keep getting the answer, 1^(infinity)=1. My thinking when working with the exponent, (1/h); as h approaches 0, the overall exponent value will near infinity. When working with the base, (1-2h), my reasoning is that 1-(a very small number close to 0), so the overall base value is 1. That is how I get 1^infinity. I checked the answer and it is different than mine, so I am confused.

Thanks in advance for any hints or help you can provide!

lim (1-2h)^(1/h)

take logs
lim log = 1/h log(1-2h) = log(1-2h)/h
L'Hospital:
lim log = [-2/(1-2h)]/1 = -2
so, lim = e^-2

This makes sense, since

lim (1+ax)^(1/x) = e^ax
and x = -2

For details, visit wolframalpha.com and enter

limit (1-2x)^(1/x) as x->0

and click the Step-by-Step Solution button

To determine the limit as h approaches 0 of the expression (1-2h)^(1/h), we can use the concept of exponential limit laws. Let's go step by step:

1. Start by taking the natural logarithm (ln) of both sides of the expression:
ln((1-2h)^(1/h))

Applying the laws of logarithms, we can rewrite this as:
ln(1-2h) / h

2. Now, we can apply L'Hôpital's Rule. According to this rule, if we have an indeterminate form of the type 0/0 or infinity/infinity, we can take the derivative of the numerator and denominator until we get a limit that is no longer indeterminate.

Taking the derivative of ln(1-2h), we get -2/(1-2h).

Taking the derivative of h, we get 1.

Applying L'Hôpital's Rule, we have:
lim(h->0) [-2/(1-2h)] / 1

3. Simplifying the expression, we have:
lim(h->0) [-2/(1-2h)]

4. Now, substitute h=0 into the expression:
-2/(1-2(0)) = -2/1 = -2

Therefore, the limit as h approaches 0 of (1-2h)^(1/h) is -2.

Your earlier reasoning of 1^infinity is incorrect because 1 raised to the power of any value (including infinity) is always equal to 1. However, in this case, we need to apply the logarithm and use L'Hôpital's Rule to evaluate the limit properly.