Answer :
To find the limit \(\lim_{x \to \frac{1}{\sqrt{2}}} \frac{4x^2 - 2}{8x^3 - \sqrt{8}}\), let's go through the solution step by step.
1. Substitute \( x = \frac{1}{\sqrt{2}} \) directly into the function:
First, substitute \( x \to \frac{1}{\sqrt{2}} \) directly into the numerator and the denominator to check if the function is indeterminate.
[tex]\[ \text{Numerator: } 4\left(\frac{1}{\sqrt{2}}\right)^2 - 2 = 4 \left( \frac{1}{2} \right) - 2 = 2 - 2 = 0 \][/tex]
[tex]\[ \text{Denominator: } 8\left(\frac{1}{\sqrt{2}}\right)^3 - \sqrt{8} = 8 \left( \frac{1}{2\sqrt{2}} \right) - \sqrt{8} = 8 \left( \frac{1}{2\sqrt{2}} \right) - 2\sqrt{2} = 4 \left( \frac{1}{\sqrt{2}} \right) - 2\sqrt{2} = 2\sqrt{2} - 2\sqrt{2} = 0 \][/tex]
Since both the numerator and the denominator are zero when \( x = \frac{1}{\sqrt{2}} \), we have an indeterminate form \( \frac{0}{0} \). Therefore, we need to apply L'Hôpital's rule.
2. Apply L'Hôpital's Rule:
According to L'Hôpital's Rule, if you have an indeterminate form of type \( \frac{0}{0} \), you can take the derivative of the numerator and the denominator and then take the limit again.
Let's differentiate the numerator and the denominator separately:
- Numerator: \( 4x^2 - 2 \)
[tex]\[ \text{Derivative of the numerator: } \frac{d}{dx} (4x^2 - 2) = 8x \][/tex]
- Denominator: \( 8x^3 - \sqrt{8} \)
[tex]\[ \text{Derivative of the denominator: } \frac{d}{dx} (8x^3 - \sqrt{8}) = 24x^2 \][/tex]
3. Find the new limit using the derivatives:
Now we take the limit of the new function created by taking these derivatives:
[tex]\[ \lim_{x \to \frac{1}{\sqrt{2}}} \frac{8x}{24x^2} \][/tex]
4. Simplify the expression:
Simplify the fraction:
[tex]\[ \frac{8x}{24x^2} = \frac{8}{24x} = \frac{1}{3x} \][/tex]
5. Substitute \( x = \frac{1}{\sqrt{2}} \) into the simplified expression:
[tex]\[ \lim_{x \to \frac{1}{\sqrt{2}}} \frac{1}{3x} = \frac{1}{3 \cdot \frac{1}{\sqrt{2}}} = \frac{1}{\frac{3}{\sqrt{2}}} = \frac{\sqrt{2}}{3} \][/tex]
Therefore, the limit is
[tex]\[ \boxed{\frac{\sqrt{2}}{3}} \][/tex]
1. Substitute \( x = \frac{1}{\sqrt{2}} \) directly into the function:
First, substitute \( x \to \frac{1}{\sqrt{2}} \) directly into the numerator and the denominator to check if the function is indeterminate.
[tex]\[ \text{Numerator: } 4\left(\frac{1}{\sqrt{2}}\right)^2 - 2 = 4 \left( \frac{1}{2} \right) - 2 = 2 - 2 = 0 \][/tex]
[tex]\[ \text{Denominator: } 8\left(\frac{1}{\sqrt{2}}\right)^3 - \sqrt{8} = 8 \left( \frac{1}{2\sqrt{2}} \right) - \sqrt{8} = 8 \left( \frac{1}{2\sqrt{2}} \right) - 2\sqrt{2} = 4 \left( \frac{1}{\sqrt{2}} \right) - 2\sqrt{2} = 2\sqrt{2} - 2\sqrt{2} = 0 \][/tex]
Since both the numerator and the denominator are zero when \( x = \frac{1}{\sqrt{2}} \), we have an indeterminate form \( \frac{0}{0} \). Therefore, we need to apply L'Hôpital's rule.
2. Apply L'Hôpital's Rule:
According to L'Hôpital's Rule, if you have an indeterminate form of type \( \frac{0}{0} \), you can take the derivative of the numerator and the denominator and then take the limit again.
Let's differentiate the numerator and the denominator separately:
- Numerator: \( 4x^2 - 2 \)
[tex]\[ \text{Derivative of the numerator: } \frac{d}{dx} (4x^2 - 2) = 8x \][/tex]
- Denominator: \( 8x^3 - \sqrt{8} \)
[tex]\[ \text{Derivative of the denominator: } \frac{d}{dx} (8x^3 - \sqrt{8}) = 24x^2 \][/tex]
3. Find the new limit using the derivatives:
Now we take the limit of the new function created by taking these derivatives:
[tex]\[ \lim_{x \to \frac{1}{\sqrt{2}}} \frac{8x}{24x^2} \][/tex]
4. Simplify the expression:
Simplify the fraction:
[tex]\[ \frac{8x}{24x^2} = \frac{8}{24x} = \frac{1}{3x} \][/tex]
5. Substitute \( x = \frac{1}{\sqrt{2}} \) into the simplified expression:
[tex]\[ \lim_{x \to \frac{1}{\sqrt{2}}} \frac{1}{3x} = \frac{1}{3 \cdot \frac{1}{\sqrt{2}}} = \frac{1}{\frac{3}{\sqrt{2}}} = \frac{\sqrt{2}}{3} \][/tex]
Therefore, the limit is
[tex]\[ \boxed{\frac{\sqrt{2}}{3}} \][/tex]
