To evaluate the given limits as [tex]\( x \to \infty \)[/tex], let's analyze each expression step by step.
## For the first expression: [tex]\(\frac{x}{\sqrt[3]{x}}\)[/tex]
1. The cube root of [tex]\( x \)[/tex] (written as [tex]\( x^{1/3} \)[/tex]) is the same as [tex]\( x \)[/tex] raised to the power of [tex]\( \frac{1}{3} \)[/tex].
2. Thus, the given expression can be rewritten as:
[tex]\[
\frac{x}{x^{1/3}}
\][/tex]
3. Recall that when you divide powers with the same base, you subtract the exponents:
[tex]\[
\frac{x^1}{x^{1/3}} = x^{1 - 1/3} = x^{2/3}
\][/tex]
4. To find the limit as [tex]\( x \)[/tex] approaches infinity:
[tex]\[
\lim_{{x \to \infty}} x^{2/3}
\][/tex]
Since [tex]\( x^{2/3} \)[/tex] grows without bound as [tex]\( x \)[/tex] increases, the limit is:
[tex]\[
\lim_{{x \to \infty}} x^{2/3} = \infty
\][/tex]
## For the second expression: [tex]\( x + \frac{\sqrt[3]{x}}{x + \sqrt[3]{x}} \)[/tex]
1. The cube root of [tex]\( x \)[/tex] can again be written as [tex]\( x^{1/3} \)[/tex].
2. Substitute this into the given expression:
[tex]\[
x + \frac{x^{1/3}}{x + x^{1/3}}
\][/tex]
3. Divide both the numerator and the denominator of the fraction by [tex]\( x \)[/tex]:
[tex]\[
x + \frac{\frac{x^{1/3}}{x}}{\frac{x}{x} + \frac{x^{1/3}}{x}} = x + \frac{x^{1/3 - 1}}{1 + x^{1/3 - 1}} = x + \frac{x^{-2/3}}{1 + x^{-2/3}}
\][/tex]
4. As [tex]\( x \)[/tex] approaches infinity, [tex]\( x^{-2/3} \)[/tex] (which is the same as [tex]\( \frac{1}{x^{2/3}} \)[/tex]) approaches 0:
[tex]\[
x + \frac{0}{1 + 0} = x + 0 = x
\][/tex]
5. Therefore:
[tex]\[
\lim_{{x \to \infty}} \left( x + \frac{x^{1/3}}{x + x^{1/3}} \right) = \infty
\][/tex]
So, the limits of the given expressions as [tex]\( x \)[/tex] approaches infinity are both infinity:
[tex]\[
\lim_{{x \to \infty}} \frac{x}{\sqrt[3]{x}} = \infty
\][/tex]
[tex]\[
\lim_{{x \to \infty}} \left( x + \frac{\sqrt[3]{x}}{x + \sqrt[3]{x}} \right) = \infty
\][/tex]
