## lim x->∞ (1+x)^(1/x)

### To evaluate the limit as x approaches infinity of (1 + x)^(1/x), we can use the properties of exponential and logarithmic functions

To evaluate the limit as x approaches infinity of (1 + x)^(1/x), we can use the properties of exponential and logarithmic functions.

First, let’s rewrite the expression using the definition of logarithms. Taking the natural logarithm of both sides:

ln[(1 + x)^(1/x)] = ln[(1 + x)] / x

Next, we can simplify the expression using logarithmic properties. Applying the quotient property of logarithms:

ln[(1 + x)] – ln[x]

Now, as x approaches infinity, ln[x] will also approach infinity. This means the second term ln[x] becomes negligible in the overall expression. Therefore, we can ignore it when we take the limit:

lim x->∞ (ln[(1 + x)] – ln[x])

Now, let’s simplify further by applying the properties of logarithms. Using the property ln(a) – ln(b) = ln(a/b):

ln[(1 + x) / x]

Now, we can write the limit expression as:

lim x->∞ ln[(1 + x) / x]

To evaluate this limit, we can use L’Hôpital’s Rule. This rule states that if we have a limit that is in the form of 0/0 or ∞/∞, and if taking the derivative of both the numerator and denominator separately still yields the same form, then the limit can be found by taking the limit of the derivatives.

Let’s differentiate the numerator and denominator separately:

Numerator: d/dx (1 + x) = 1

Denominator: d/dx (x) = 1

So, applying L’Hôpital’s Rule:

lim x->∞ ln[(1 + x) / x] = lim x->∞ (1/x) / 1

Taking the limit as x approaches infinity:

lim x->∞ (1/x) / 1 = 0

Therefore, the limit of (1 + x)^(1/x) as x approaches infinity is 0.

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