Key Concepts and Formulas

• Limits and Indeterminate Forms: We will encounter the indeterminate form $\frac{0}{0}$, which suggests the application of L'Hôpital's Rule or algebraic manipulation.
• L'Hôpital's Rule: For functions $f(x)$ and $g(x)$ such that $\lim_{x \to c} f(x) = 0$ and $\lim_{x \to c} g(x) = 0$, then $\lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)}$, provided the latter limit exists.
• Logarithmic Differentiation/Power Rule for Differentiation: The derivative of $u(x)^n$ is $n \cdot u(x)^{n-1} \cdot u'(x)$.
• Rationalization (for roots): Multiplying by the conjugate can help simplify expressions involving roots, although L'Hôpital's rule is often more direct for limits.

Step-by-Step Solution

Step 1: Identify the Indeterminate Form We are asked to evaluate the limit: I=\lim _\limits{x \rightarrow 1} \frac{(5 x+1)^{1 / 3}-(x+5)^{1 / 3}}{(2 x+3)^{1 / 2}-(x+4)^{1 / 2}} Let's substitute $x=1$ into the numerator and the denominator: Numerator: $(5(1)+1)^{1/3} - (1+5)^{1/3} = 6^{1/3} - 6^{1/3} = 0$ Denominator: $(2(1)+3)^{1/2} - (1+4)^{1/2} = 5^{1/2} - 5^{1/2} = 0$ Since we have the indeterminate form $\frac{0}{0}$, we can apply L'Hôpital's Rule.

Step 2: Apply L'Hôpital's Rule We need to find the derivatives of the numerator and the denominator with respect to $x$. Let $f(x) = (5x+1)^{1/3} - (x+5)^{1/3}$ and $g(x) = (2x+3)^{1/2} - (x+4)^{1/2}$. Using the power rule for differentiation $(u^n)' = n u^{n-1} u'$: $f'(x) = \frac{1}{3}(5x+1)^{-2/3} \cdot 5 - \frac{1}{3}(x+5)^{-2/3} \cdot 1$ $g'(x) = \frac{1}{2}(2x+3)^{-1/2} \cdot 2 - \frac{1}{2}(x+4)^{-1/2} \cdot 1$

Now, apply L'Hôpital's Rule: $I = \lim _{x \rightarrow 1} \frac{f'(x)}{g'(x)} = \lim _{x \rightarrow 1} \frac{\frac{5}{3}(5 x+1)^{-2 / 3}-\frac{1}{3}(x+5)^{-2 / 3}}{\frac{1}{2}(2 x+3)^{-1 / 2}-\frac{1}{2}(x+4)^{-1 / 2}}$

Step 3: Evaluate the Limit by Substitution Substitute $x=1$ into the expression for $\frac{f'(x)}{g'(x)}$: $I = \frac{\frac{5}{3}(5(1)+1)^{-2 / 3}-\frac{1}{3}(1+5)^{-2 / 3}}{\frac{1}{2}(2(1)+3)^{-1 / 2}-\frac{1}{2}(1+4)^{-1 / 2}}$ $I = \frac{\frac{5}{3}(6)^{-2 / 3}-\frac{1}{3}(6)^{-2 / 3}}{\frac{1}{2}(5)^{-1 / 2}-\frac{1}{2}(5)^{-1 / 2}}$ The denominator becomes $0$, which means L'Hopital's rule was applied incorrectly in the original solution. Let's re-evaluate the derivatives.

Correction to Step 2 and Step 3:

Step 2 (Corrected): Apply L'Hôpital's Rule Let $f(x) = (5x+1)^{1/3} - (x+5)^{1/3}$ and $g(x) = (2x+3)^{1/2} - (x+4)^{1/2}$. $f'(x) = \frac{d}{dx} [(5x+1)^{1/3}] - \frac{d}{dx} [(x+5)^{1/3}]$ $f'(x) = \frac{1}{3}(5x+1)^{-2/3} \cdot 5 - \frac{1}{3}(x+5)^{-2/3} \cdot 1$ $f'(x) = \frac{5}{3}(5x+1)^{-2/3} - \frac{1}{3}(x+5)^{-2/3}$

$g'(x) = \frac{d}{dx} [(2x+3)^{1/2}] - \frac{d}{dx} [(x+4)^{1/2}]$ $g'(x) = \frac{1}{2}(2x+3)^{-1/2} \cdot 2 - \frac{1}{2}(x+4)^{-1/2} \cdot 1$ $g'(x) = (2x+3)^{-1/2} - \frac{1}{2}(x+4)^{-1/2}$

Now, apply L'Hôpital's Rule: $I = \lim _{x \rightarrow 1} \frac{f'(x)}{g'(x)} = \lim _{x \rightarrow 1} \frac{\frac{5}{3}(5 x+1)^{-2 / 3}-\frac{1}{3}(x+5)^{-2 / 3}}{(2 x+3)^{-1 / 2}-\frac{1}{2}(x+4)^{-1 / 2}}$

Step 3 (Corrected): Evaluate the Limit by Substitution Substitute $x=1$ into the expression for $\frac{f'(x)}{g'(x)}$: $I = \frac{\frac{5}{3}(5(1)+1)^{-2 / 3}-\frac{1}{3}(1+5)^{-2 / 3}}{(2(1)+3)^{-1 / 2}-\frac{1}{2}(1+4)^{-1 / 2}}$ $I = \frac{\frac{5}{3}(6)^{-2 / 3}-\frac{1}{3}(6)^{-2 / 3}}{(5)^{-1 / 2}-\frac{1}{2}(5)^{-1 / 2}}$ $I = \frac{\left(\frac{5}{3}-\frac{1}{3}\right)(6)^{-2 / 3}}{\left(1-\frac{1}{2}\right)(5)^{-1 / 2}}$ $I = \frac{\frac{4}{3}(6)^{-2 / 3}}{\frac{1}{2}(5)^{-1 / 2}}$ $I = \frac{4}{3} \cdot \frac{2}{1} \cdot \frac{(5)^{-1 / 2}}{(6)^{2 / 3}}$ $I = \frac{8}{3} \cdot \frac{5^{1 / 2}}{6^{2 / 3}}$

Step 4: Match the Result with the Given Form We are given that the limit is equal to $\frac{\mathrm{m} \sqrt{5}}{\mathrm{n}(2 \mathrm{n})^{2 / 3}}$. Our calculated limit is $I = \frac{8}{3} \cdot \frac{\sqrt{5}}{6^{2 / 3}}$. We can rewrite $6^{2/3}$ as $(3 \cdot 2)^{2/3} = 3^{2/3} \cdot 2^{2/3}$. This doesn't seem to directly match the structure.

Let's re-examine the given form: $\frac{\mathrm{m} \sqrt{5}}{\mathrm{n}(2 \mathrm{n})^{2 / 3}}$. Let's try to manipulate our result to match this form. $I = \frac{8 \sqrt{5}}{3 \cdot 6^{2/3}}$ We need the denominator to be in the form $n(2n)^{2/3}$. If we let $n=3$, then $n(2n)^{2/3} = 3(2 \cdot 3)^{2/3} = 3(6)^{2/3}$. This matches the denominator in our expression. So, by comparing $I = \frac{8 \sqrt{5}}{3 \cdot 6^{2 / 3}}$ with $\frac{\mathrm{m} \sqrt{5}}{\mathrm{n}(2 \mathrm{n})^{2 / 3}}$, we can identify: $m = 8$ $n = 3$

Let's verify the gcd condition: $\operatorname{gcd}(m, n) = \operatorname{gcd}(8, 3) = 1$. This condition is satisfied.

Step 5: Calculate the Final Expression We need to find the value of $8m + 12n$. Substitute the values of $m=8$ and $n=3$: $8m + 12n = 8(8) + 12(3)$ $8m + 12n = 64 + 36$ $8m + 12n = 100$

Common Mistakes & Tips

• Incorrect application of L'Hôpital's Rule: Ensure the indeterminate form $\frac{0}{0}$ or $\frac{\infty}{\infty}$ is present before applying the rule. Also, differentiate the numerator and denominator separately, not as a quotient.
• Algebraic errors with exponents: Be careful when manipulating terms with fractional and negative exponents. For instance, $(a^b)^c = a^{bc}$ and $a^{-b} = \frac{1}{a^b}$.
• Matching the final form: After calculating the limit, carefully compare it with the given symbolic form to correctly identify the values of $m$ and $n$. Ensure all parts of the given form are accounted for.

Summary

The problem involves evaluating a limit that results in an indeterminate form $\frac{0}{0}$. L'Hôpital's Rule is applied by differentiating the numerator and the denominator separately. After correctly calculating the derivatives and substituting $x=1$, the limit is found to be $\frac{8}{3} \frac{\sqrt{5}}{6^{2/3}}$. By comparing this result with the given form $\frac{\mathrm{m} \sqrt{5}}{\mathrm{n}(2 \mathrm{n})^{2 / 3}}$, we identify $m=8$ and $n=3$, satisfying $\operatorname{gcd}(m, n)=1$. Finally, the expression $8m + 12n$ is calculated to be $100$.

The final answer is $\boxed{100}$.