Exploring the Limit of the Function √(x^2 3) / (x - 7) as x Approaches 7
In this article, we will explore the limit of the function (frac{sqrt{x^2 3}}{x - 7}) as (x) approaches 7. We will apply L'H?pital's Rule and discuss the conditions under which the limit exists.
Initial Analysis
When we substitute (x 7) into the function, we obtain:
(frac{sqrt{7^2 3}}{7 - 7} frac{sqrt{52}}{0})This leads to an indeterminate form of the type (frac{k}{0}), where (k) is a finite number. Therefore, we need to investigate further to determine if the limit exists.
Applying L'H?pital's Rule
L'H?pital's Rule is applicable to limits of the form (frac{0}{0}) or (frac{infty}{infty}). For the given function, substituting (x 7) gives us an indeterminate form of (frac{0}{0}). Therefore, we can apply L'H?pital's Rule by taking the derivatives of the numerator and the denominator.
Numerator
The numerator is (sqrt{x^2 3}). To find its derivative, we use the chain rule:
(frac{d}{dx}left(sqrt{x^2 3}right) frac{1}{2sqrt{x^2 3}} cdot 2x frac{x}{sqrt{x^2 3}})Denominator
The denominator is (x - 7). Its derivative is:
(frac{d}{dx}(x - 7) 1)Applying the Rule
Now, we can apply L'H?pital's Rule:
(lim_{x to 7} frac{sqrt{x^2 3}}{x - 7} lim_{x to 7} frac{frac{x}{sqrt{x^2 3}}}{1} lim_{x to 7} frac{x}{sqrt{x^2 3}})Substituting (x 7) into the expression, we get:
(frac{7}{sqrt{7^2 3}} frac{7}{sqrt{52}} frac{7}{2sqrt{13}} frac{7}{2 cdot sqrt{13}} frac{7}{2 cdot 3.6056} approx 0.9722)This simplifies to:
(frac{7}{2 cdot 3.6056} frac{7}{7.2112} approx 0.9722)However, this is not the correct approach for our function. We need to consider the form of the function again and correct any mistakes.
Correcting the Approach
Upon careful examination, it appears that there might be a mistake in the initial application of L'H?pital's Rule. Let's correct the approach:
Revised Derivative Calculation
The correct derivative of the numerator should be:
(frac{d}{dx}left(sqrt{x^2 3}right) frac{x}{sqrt{x^2 3}})Thus, applying L'H?pital's Rule:
(lim_{x to 7} frac{sqrt{x^2 3}}{x - 7} lim_{x to 7} frac{frac{x}{sqrt{x^2 3}}}{1} lim_{x to 7} frac{x}{sqrt{x^2 3}})Substituting (x 7) into this expression, we get:
(frac{7}{sqrt{52}} frac{7}{2sqrt{13}} frac{7}{2 cdot 3.6056} frac{7}{7.2112} approx 0.9722)However, we need to re-evaluate the function more carefully.
Revisiting the Original Function
Upon reviewing the original function, (frac{sqrt{x^2 3}}{x - 7}), we realize it is not directly solvable using L'H?pital's Rule in the given form. Instead, we need to simplify it further.
Alternative Approach
Let's rewrite the function as:
(frac{sqrt{x^2 3}}{x - 7} frac{sqrt{x^2 3} - 3 3}{x - 7} frac{sqrt{x^2 3} - 3}{x - 7} frac{3}{x - 7})Since the second term (frac{3}{x - 7}) clearly goes to infinity as (x to 7), we can focus on the first term:
(lim_{x to 7} frac{sqrt{x^2 3} - 3}{x - 7})This simplifies the problem. By applying L'H?pital's Rule again:
(lim_{x to 7} frac{sqrt{x^2 3} - 3}{x - 7} lim_{x to 7} frac{frac{x}{sqrt{x^2 3}}}{1} frac{7}{sqrt{52}} frac{7}{2sqrt{13}} frac{7}{2 cdot 3.6056} approx 0.9722)Thus, the limit of the first term is:
(frac{7}{2sqrt{13}} frac{7}{2 cdot 3.6056} frac{7}{7.2112} approx 0.9722)The second term clearly tends to infinity, so the overall limit does not exist.
Conclusion
The limit of the function (frac{sqrt{x^2 3}}{x - 7}) as (x) approaches 7 is not clearly defined and typically does not exist due to the behavior of the denominator approaching zero while the numerator approaches a finite value.
It is important to note that for the function (frac{sqrt{x^2 - 3}}{x - 7}), the limit can be evaluated as:
(lim_{x to 7} frac{sqrt{x^2 - 3}}{x - 7} frac{1}{2sqrt{13}} frac{1}{6})This is derived from the derivative of the numerator and denominator, as outlined earlier.