Answer :
To find the limit [tex]\(\lim_{{x \to a}} \frac{x^{\frac{1}{3}} - a^{\frac{1}{3}}}{x^{\frac{1}{2}} - a^{\frac{1}{2}}}\)[/tex], we can proceed with the following steps:
1. Expression Analysis and Substitution:
We start with the original limit:
[tex]\[ \lim_{{x \to a}} \frac{x^{\frac{1}{3}} - a^{\frac{1}{3}}}{x^{\frac{1}{2}} - a^{\frac{1}{2}}} \][/tex]
Direct substitution [tex]\(x = a\)[/tex] leads to [tex]\( \frac{0}{0} \)[/tex], which is an indeterminate form. Hence, we need to simplify or rework the expression to resolve this indeterminacy.
2. Change of Variables:
Consider the substitution [tex]\(u = x^{\frac{1}{6}}\)[/tex]. Therefore, [tex]\(x = u^6\)[/tex] and similarly for [tex]\(a\)[/tex], let [tex]\(v = a^{\frac{1}{6}}\)[/tex], then [tex]\(a = v^6\)[/tex]. Thus, our limit now transforms:
[tex]\[ \lim_{{u \to v}} \frac{(u^6)^{\frac{1}{3}} - (v^6)^{\frac{1}{3}}}{(u^6)^{\frac{1}{2}} - (v^6)^{\frac{1}{2}}} \][/tex]
3. Simplify the New Expressions:
Simplify the exponents:
[tex]\[ (u^6)^{\frac{1}{3}} = u^2 \quad \text{and} \quad (v^6)^{\frac{1}{3}} = v^2 \][/tex]
[tex]\[ (u^6)^{\frac{1}{2}} = u^3 \quad \text{and} \quad (v^6)^{\frac{1}{2}} = v^3 \][/tex]
So, the limit now becomes:
[tex]\[ \lim_{{u \to v}} \frac{u^2 - v^2}{u^3 - v^3} \][/tex]
4. Factorizing the Simplified Fractions:
Use the difference of squares and cubes formulas:
[tex]\[ u^2 - v^2 = (u - v)(u + v) \][/tex]
[tex]\[ u^3 - v^3 = (u - v)(u^2 + uv + v^2) \][/tex]
The fraction simplifies to:
[tex]\[ \frac{u^2 - v^2}{u^3 - v^3} = \frac{(u - v)(u + v)}{(u - v)(u^2 + uv + v^2)} \][/tex]
5. Cancel Common Factors:
Cancel the common factor [tex]\((u - v)\)[/tex] from numerator and denominator:
[tex]\[ \frac{(u - v)(u + v)}{(u - v)(u^2 + uv + v^2)} = \frac{u + v}{u^2 + uv + v^2} \][/tex]
6. Taking the Limit:
With the simplified fraction:
[tex]\[ \lim_{{u \to v}} \frac{u + v}{u^2 + uv + v^2} \][/tex]
We substitute [tex]\(u = v\)[/tex] into the simplified form:
[tex]\[ \frac{v + v}{v^2 + vv + v^2} = \frac{2v}{3v^2} = \frac{2}{3v} = \frac{2}{3a^{\frac{1}{6}}} \][/tex]
Therefore, the limit is:
[tex]\[ \boxed{\frac{2}{3a^{\frac{1}{6}}}} \][/tex]
1. Expression Analysis and Substitution:
We start with the original limit:
[tex]\[ \lim_{{x \to a}} \frac{x^{\frac{1}{3}} - a^{\frac{1}{3}}}{x^{\frac{1}{2}} - a^{\frac{1}{2}}} \][/tex]
Direct substitution [tex]\(x = a\)[/tex] leads to [tex]\( \frac{0}{0} \)[/tex], which is an indeterminate form. Hence, we need to simplify or rework the expression to resolve this indeterminacy.
2. Change of Variables:
Consider the substitution [tex]\(u = x^{\frac{1}{6}}\)[/tex]. Therefore, [tex]\(x = u^6\)[/tex] and similarly for [tex]\(a\)[/tex], let [tex]\(v = a^{\frac{1}{6}}\)[/tex], then [tex]\(a = v^6\)[/tex]. Thus, our limit now transforms:
[tex]\[ \lim_{{u \to v}} \frac{(u^6)^{\frac{1}{3}} - (v^6)^{\frac{1}{3}}}{(u^6)^{\frac{1}{2}} - (v^6)^{\frac{1}{2}}} \][/tex]
3. Simplify the New Expressions:
Simplify the exponents:
[tex]\[ (u^6)^{\frac{1}{3}} = u^2 \quad \text{and} \quad (v^6)^{\frac{1}{3}} = v^2 \][/tex]
[tex]\[ (u^6)^{\frac{1}{2}} = u^3 \quad \text{and} \quad (v^6)^{\frac{1}{2}} = v^3 \][/tex]
So, the limit now becomes:
[tex]\[ \lim_{{u \to v}} \frac{u^2 - v^2}{u^3 - v^3} \][/tex]
4. Factorizing the Simplified Fractions:
Use the difference of squares and cubes formulas:
[tex]\[ u^2 - v^2 = (u - v)(u + v) \][/tex]
[tex]\[ u^3 - v^3 = (u - v)(u^2 + uv + v^2) \][/tex]
The fraction simplifies to:
[tex]\[ \frac{u^2 - v^2}{u^3 - v^3} = \frac{(u - v)(u + v)}{(u - v)(u^2 + uv + v^2)} \][/tex]
5. Cancel Common Factors:
Cancel the common factor [tex]\((u - v)\)[/tex] from numerator and denominator:
[tex]\[ \frac{(u - v)(u + v)}{(u - v)(u^2 + uv + v^2)} = \frac{u + v}{u^2 + uv + v^2} \][/tex]
6. Taking the Limit:
With the simplified fraction:
[tex]\[ \lim_{{u \to v}} \frac{u + v}{u^2 + uv + v^2} \][/tex]
We substitute [tex]\(u = v\)[/tex] into the simplified form:
[tex]\[ \frac{v + v}{v^2 + vv + v^2} = \frac{2v}{3v^2} = \frac{2}{3v} = \frac{2}{3a^{\frac{1}{6}}} \][/tex]
Therefore, the limit is:
[tex]\[ \boxed{\frac{2}{3a^{\frac{1}{6}}}} \][/tex]