Math Problem Statement

limπ‘₯β†’232(4π‘₯2+8π‘₯βˆ’6)3/2

Solution

To evaluate the limit

lim⁑xβ†’232(4x2+8xβˆ’6)3/2,\lim_{x \to 232} \left(4x^2 + 8x - 6\right)^{3/2},

we can directly substitute x=232x = 232 into the expression because the function inside the limit is continuous. Let's compute this step by step.

Step 1: Substitute x=232x = 232 into the quadratic expression

4x2+8xβˆ’6=4(232)2+8(232)βˆ’64x^2 + 8x - 6 = 4(232)^2 + 8(232) - 6

Step 2: Compute 2322232^2

2322=53824232^2 = 53824

Step 3: Multiply by 4

4Γ—53824=2152964 \times 53824 = 215296

Step 4: Compute 8Γ—2328 \times 232

8Γ—232=18568 \times 232 = 1856

Step 5: Add the results and subtract 6

215296+1856βˆ’6=217146215296 + 1856 - 6 = 217146

Step 6: Raise the result to the power of 3/23/2

Final result=(217146)3/2\text{Final result} = (217146)^{3/2}

Step 7: Compute 2171463/2217146^{3/2}

This is equivalent to taking the square root of 217146 and then cubing the result:

217146β‰ˆ466(approximation)\sqrt{217146} \approx 466 \quad \text{(approximation)}

Then:

4663=101194696466^3 = 101194696

So,

lim⁑xβ†’232(4x2+8xβˆ’6)3/2β‰ˆ101194696.\lim_{x \to 232} \left(4x^2 + 8x - 6\right)^{3/2} \approx 101194696.

This is the value of the limit.


Would you like more details, or do you have any questions? Here are some related questions you might find useful:

  1. What is the meaning of continuity in the context of limits?
  2. How do you handle limits that approach infinity or negative infinity?
  3. What are some techniques for evaluating limits without direct substitution?
  4. How does the power rule work when dealing with limits?
  5. What is the significance of the exponent 32\frac{3}{2} in this context?

Tip: When dealing with limits, always check if the function is continuous at the point of interest, as this allows for direct substitution to find the limit.

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Math Problem Analysis

Mathematical Concepts

Limits
Continuity
Exponentiation

Formulas

Limit definition

Theorems

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Suitable Grade Level

Advanced High School