Answer :
To find the limit \(\lim _{x \rightarrow 8} \frac{\sqrt{7+\sqrt[3]{x}}-3}{x-8}\), we will use the method of substitution and algebraic manipulation to solve this step-by-step.
First, observe that as \(x\) approaches 8, both the numerator and denominator approach 0, resulting in an indeterminate form of \(\frac{0}{0}\). We need to simplify this expression to resolve the indeterminate form.
Let's define a substitution to make the expression simpler:
Let \(u = \sqrt[3]{x}\). Consequently, \(u^3 = x\). As \(x \to 8\), we have \(u \to \sqrt[3]{8} = 2\).
Rewriting the limit in terms of \(u\):
[tex]\[ \lim_{x \to 8} \frac{\sqrt{7+\sqrt[3]{x}}-3}{x-8} = \lim_{u \to 2} \frac{\sqrt{7+u}-3}{u^3 - 8} \][/tex]
Notice that \(u^3 - 8\) can be factored as \((u - 2)(u^2 + 2u + 4)\):
[tex]\[ \lim_{u \to 2} \frac{\sqrt{7 + u} - 3}{(u - 2)(u^2 + 2u + 4)} \][/tex]
To eliminate the indeterminate form, we will rationalize the numerator. Multiply the numerator and the denominator by the conjugate of the numerator \(\sqrt{7 + u} + 3\):
[tex]\[ \lim_{u \to 2} \frac{(\sqrt{7 + u} - 3)(\sqrt{7 + u} + 3)}{(u - 2)(u^2 + 2u + 4)(\sqrt{7 + u} + 3)} \][/tex]
In the numerator, \((\sqrt{7 + u} - 3)(\sqrt{7 + u} + 3)\) simplifies to:
[tex]\[ (\sqrt{7 + u})^2 - 3^2 = 7 + u - 9 = u - 2 \][/tex]
So the limit becomes:
[tex]\[ \lim_{u \to 2} \frac{u - 2}{(u - 2)(u^2 + 2u + 4)(\sqrt{7 + u} + 3)} \][/tex]
Now, cancel the common factor \(u - 2\):
[tex]\[ \lim_{u \to 2} \frac{1}{(u^2 + 2u + 4)(\sqrt{7 + u} + 3)} \][/tex]
Finally substitute \(u = 2\):
[tex]\[ \frac{1}{(2^2 + 2 \cdot 2 + 4)(\sqrt{7 + 2} + 3)} = \frac{1}{(4 + 4 + 4)(\sqrt{9} + 3)} = \frac{1}{(12)(3 + 3)} = \frac{1}{12 \cdot 6} = \frac{1}{72} \][/tex]
Therefore, the limit is:
[tex]\[ \boxed{\frac{1}{72}} \][/tex]
First, observe that as \(x\) approaches 8, both the numerator and denominator approach 0, resulting in an indeterminate form of \(\frac{0}{0}\). We need to simplify this expression to resolve the indeterminate form.
Let's define a substitution to make the expression simpler:
Let \(u = \sqrt[3]{x}\). Consequently, \(u^3 = x\). As \(x \to 8\), we have \(u \to \sqrt[3]{8} = 2\).
Rewriting the limit in terms of \(u\):
[tex]\[ \lim_{x \to 8} \frac{\sqrt{7+\sqrt[3]{x}}-3}{x-8} = \lim_{u \to 2} \frac{\sqrt{7+u}-3}{u^3 - 8} \][/tex]
Notice that \(u^3 - 8\) can be factored as \((u - 2)(u^2 + 2u + 4)\):
[tex]\[ \lim_{u \to 2} \frac{\sqrt{7 + u} - 3}{(u - 2)(u^2 + 2u + 4)} \][/tex]
To eliminate the indeterminate form, we will rationalize the numerator. Multiply the numerator and the denominator by the conjugate of the numerator \(\sqrt{7 + u} + 3\):
[tex]\[ \lim_{u \to 2} \frac{(\sqrt{7 + u} - 3)(\sqrt{7 + u} + 3)}{(u - 2)(u^2 + 2u + 4)(\sqrt{7 + u} + 3)} \][/tex]
In the numerator, \((\sqrt{7 + u} - 3)(\sqrt{7 + u} + 3)\) simplifies to:
[tex]\[ (\sqrt{7 + u})^2 - 3^2 = 7 + u - 9 = u - 2 \][/tex]
So the limit becomes:
[tex]\[ \lim_{u \to 2} \frac{u - 2}{(u - 2)(u^2 + 2u + 4)(\sqrt{7 + u} + 3)} \][/tex]
Now, cancel the common factor \(u - 2\):
[tex]\[ \lim_{u \to 2} \frac{1}{(u^2 + 2u + 4)(\sqrt{7 + u} + 3)} \][/tex]
Finally substitute \(u = 2\):
[tex]\[ \frac{1}{(2^2 + 2 \cdot 2 + 4)(\sqrt{7 + 2} + 3)} = \frac{1}{(4 + 4 + 4)(\sqrt{9} + 3)} = \frac{1}{(12)(3 + 3)} = \frac{1}{12 \cdot 6} = \frac{1}{72} \][/tex]
Therefore, the limit is:
[tex]\[ \boxed{\frac{1}{72}} \][/tex]
