Evaluate the limit using L'Hospital's rule if necessary.

lim as x goes to +infinity x^(6/x)

I will calculate the limit of the natural log instead. The limit of x^(6/x) will be e to that power of the limit of the log.

ln x^(6/x) = (6/x)ln x = lnx/(x/6)
As x-> infinity, that becomes the limit of 6/x, which is zero

If the log of the limit is zero, the limit itself is 1.

To evaluate the limit of \(\lim_{x \to +\infty} x^{\frac{6}{x}}\), we can start by recognizing that this form is of the indeterminate form \(1^{\infty}\).

To proceed with L'Hospital's rule, we need to rewrite the expression in a different form that can help us take the derivative. We can use the natural logarithm function to achieve this.

Let's define \(y = \ln(x^{\frac{6}{x}})\). Taking the logarithm allows us to bring the exponent down as a coefficient.

Taking the natural logarithm of both sides, we have:
\(\ln(y) = \ln\left(x^{\frac{6}{x}}\right)\)

Using the exponent property of logarithms, we can bring down the exponent:
\(\ln(y) = \frac{6}{x} \ln(x)\)

Now, let's differentiate both sides with respect to \(x\) using the chain rule:

\(\frac{1}{y} \cdot \frac{dy}{dx} = \frac{6}{x} \cdot \frac{1}{x} + \frac{6}{x} \cdot \ln(x) \cdot \frac{d}{dx}(x)\)

Simplifying, we get:

\(\frac{dy}{dx} = \frac{6}{x^2} + \frac{6}{x} \ln(x)\)

Now, we can find the limit of \(\frac{dy}{dx}\) as \(x\) approaches \(+\infty\):

\(\lim_{x \to +\infty} \frac{dy}{dx} = \lim_{x \to +\infty} \left(\frac{6}{x^2} + \frac{6}{x} \ln(x)\right)\)

As \(x\) approaches \(+\infty\), the first term \(\frac{6}{x^2}\) tends to 0.

For the second term \(\frac{6}{x} \ln(x)\), the logarithm grows slower than any power of \(x\), so it also tends to 0.

Therefore, the limit becomes:

\(\lim_{x \to +\infty} \frac{dy}{dx} = 0\)

Since \(\frac{dy}{dx}\) is the derivative of \(\ln(y)\), this implies that \(\ln(y)\) approaches a constant as \(x\) approaches \(+\infty\).

In other words, \(\ln\left(x^{\frac{6}{x}}\right)\) approaches a constant value as \(x\) approaches \(+\infty\).

Now, we can exponentiate both sides to recover the original expression:

\(y = x^{\frac{6}{x}}\)

Taking the exponential function of both sides gives us:

\(e^y = e^{x^{\frac{6}{x}}}\)

As \(x\) approaches \(+\infty\), the left-hand side \(e^y\) remains finite (since \(\ln(y)\) approaches a constant). Therefore, the right-hand side \(e^{x^{\frac{6}{x}}}\) also approaches a finite value.

Finally, we conclude that:

\(\lim_{x \to +\infty} x^{\frac{6}{x}}\) equals the value that \(e^{x^{\frac{6}{x}}}\) approaches as \(x\) approaches \(+\infty\).

To evaluate the limit lim as x goes to +infinity x^(6/x), we can use L'Hospital's rule.

L'Hospital's rule states that if we have an indeterminate form of the form 0/0 or ∞/∞, we can differentiate the numerator and denominator until we obtain a determinate form (a limit that can be evaluated directly).

Let's apply L'Hospital's rule:
1. Take the natural logarithm of both sides of the expression:
ln(lim as x goes to +infinity x^(6/x)) = ln(x^(6/x))

2. Now we can rewrite the expression using the logarithm rules:
ln(lim as x goes to +infinity x^(6/x)) = (6/x) * ln(x)

3. Apply L'Hospital's rule by differentiating both the numerator and denominator with respect to x:
lim as x goes to +infinity [(6/x) * ln(x)]
Differentiating the numerator by using the quotient rule:
lim as x goes to +infinity [(6/x * 1) - (6 * (1/x^2) * ln(x))]
Simplifying the expression:
lim as x goes to +infinity [(6/x) - (6 * ln(x)/x^2)]

4. Now we can evaluate the limit by taking the limit as x approaches infinity.
As x approaches infinity, the second term [(6 * ln(x)/x^2)] tends to zero because ln(x) grows much slower than x^2.
Therefore, the only term that remains is (6/x).
So, the final answer is:
lim as x goes to +infinity x^(6/x) = 1

Therefore, the limit of x^(6/x) as x approaches infinity is 1.