Question

\(\operatorname{Lt}_{x \rightarrow 5} \frac{\sqrt{x+20}-\sqrt{3 x+10}}{5-x}=\) (1) \(-\frac{2}{5}\) (2) \(\frac{2}{5}\) (3) \(\frac{1}{5}\) (4) \(\frac{-1}{5}\)

Short answer

a) 2/5 b) 1/5 c) -1/5 d) None of the above Answer: d) None of the above. Note: There might be some mistakes in the calculations or the provided options are incorrect.

Step-by-step solution

Rationalize the numerator

To rationalize the numerator, we need to multiply both the numerator and denominator by the conjugate of the numerator. The conjugate of \(\sqrt{x+20}-\sqrt{3x+10}\) is \(\sqrt{x+20}+\sqrt{3x+10}\). So, multiply numerator and denominator by \(\sqrt{x+20} + \sqrt{3x+10}\): \(\operatorname{Lt}_{x \rightarrow 5} \frac{(\sqrt{x+20}-\sqrt{3x+10})(\sqrt{x+20}+\sqrt{3x+10})}{(5-x)(\sqrt{x+20}+\sqrt{3x+10})}\).

Simplify the numerator

Now we simplify the numerator using the difference of squares formula: \(\operatorname{Lt}_{x \rightarrow 5}\frac{(x+20) - 2\sqrt{(x+20)(3x+10)} +(3x+10)}{(5-x)(\sqrt{x+20}+\sqrt{3x+10})}\).

Combine like terms

Combine terms in the numerator: \(\operatorname{Lt}_{x \rightarrow 5}\frac{4x - 2\sqrt{(x+20)(3x+10)}}{(5-x)(\sqrt{x+20}+\sqrt{3x+10})}\).

Factor out a common factor

Factor out the common factor of 2 in the numerator: \(\operatorname{Lt}_{x \rightarrow 5}\frac{2(2x - \sqrt{(x+20)(3x+10)})}{(5-x)(\sqrt{x+20}+\sqrt{3x+10})}\).

Evaluate the limit as x approaches 5

Now, plug in \(x=5\) and simplify: \(\operatorname{Lt}_{x \rightarrow 5}\frac{2(2(5) - \sqrt{(5+20)(3(5)+10)})}{(5-5)(\sqrt{5+20}+\sqrt{3(5)+10})} = \frac{2(10 - \sqrt{25(25)})}{0(\sqrt{25}+\sqrt{25})}\). The limit is undefined as the denominator equals zero.

L'Hôpital's Rule

Since \(\lim_{x\rightarrow5}\) is of the indeterminate form \(\frac{0}{0}\), we apply L'Hôpital's rule: Find the derivatives of the numerator and denominator with respect to x: \(\frac{d}{dx}(2x - \sqrt{(x+20)(3x+10)})=-\frac{d}{dx}(5-x)\). Use the rule for Limits: \(\lim_{x \rightarrow 5} \frac{\frac{d}{dx}(2x-\sqrt{(x+20)(3x+10)})}{\frac{-d}{dx}(5-x)}\). Now, take the derivatives of both numerator and denominator: \(\frac{d}{dx}(2x-\sqrt{(x+20)(3x+10)}) = 2 - \frac{1}{2\sqrt{(x+20)(3x+10)}}[\frac{d}{dx}(x+20)(3x+10)]\) And: \(\frac{-d}{dx}(5-x) = 1\). Plug in x=5 and simplify: \(\lim_{x \rightarrow 5} \frac{2-\frac{1}{2\sqrt{(5+20)(3(5)+10)}}[\frac{d}{dx}(5+20)(3(5)+10)]}{1}\). Evaluate \(\frac{2-\frac{1}{2\sqrt{25(25)}}(4(25))}{1}=\frac{2-\frac{1}{10}(100)}{1}=\frac{2-\frac{100}{10}}{1}=\frac{2-10}{1}=\boxed{-\frac{8}{5}}\). However, this option is not listed in the given answers. The L'Hôpital's rule is not applicable here. We might have done a mistake in our calculations, or the provided options are incorrect.

Key concepts

L'Hôpital's Rule

L'Hôpital's Rule is a powerful tool used to evaluate limits of indeterminate forms. It applies when your limit initially results in forms like \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\). To use L'Hôpital's Rule, take the derivative of the numerator and the derivative of the denominator separately. Then, find the limit of this new function.

• When applying L'Hôpital’s Rule, ensure the original limit results in an indeterminate form.
• If the limit is still indeterminate after applying the rule once, you can apply it again.
• Continue deriving and evaluating until you obtain a determinate form.

This rule simplifies potentially complicated expressions, making seemingly challenging limits more approachable. It is important to check conditions since the rule only works for certain types of indeterminate forms.

Rationalization in limits

Rationalization is a technique often used to simplify fractions that contain radical expressions. It's particularly helpful in limit problems where the direct substitution results in an indeterminate form. In rationalization, multiply both the numerator and the denominator by the conjugate of the numerator.

• The conjugate of \(a - b\) is \(a + b\).
• This method turns radical expressions into simpler polynomial forms using the difference of squares.
• Once rationalized, further simplification can make evaluating the limit possible.

By turning complex expressions involving square roots into easier calculations, rationalization helps work through tricky problems efficiently. It clears up indeterminacies and allows the possibility to evaluate the limit directly by substitution or further techniques.

Indeterminate forms

In calculus, indeterminate forms are expressions that do not provide enough information to determine the limit directly. Common examples include \(\frac{0}{0}\), \(\frac{\infty}{\infty}\), \(0 \times \infty\), and others. These forms occur when trying to find limits and require special techniques for resolution.

• These situations indicate that the limit exists, but additional steps are needed to find it.
• Techniques like L'Hôpital's Rule, rationalization, and algebraic manipulation help resolve these forms.
• Understanding the nature and behavior of the function can guide which technique to use.

Recognizing such forms is the first step towards applying the correct method to resolve them. This requirement for extra steps makes understanding indeterminate forms crucial for solving limit problems effectively.

Difference of squares formula

The difference of squares is a useful algebraic formula represented as \(a^2 - b^2 = (a-b)(a+b)\). It is often utilized to simplify expressions, particularly when rationalizing numerical limits involving radicals.

• When you encounter an expression like \((\sqrt{x+20} - \sqrt{3x+10})(\sqrt{x+20} + \sqrt{3x+10})\), use \(a^2 - b^2\) to simplify.
• This formula helps eliminate radical terms by converting them into squared terms, which are easier to manipulate.
• The simplification turns a problem into a more manageable form, allowing further steps like substitution to be applied.

The difference of squares formula is simple but powerful, turning potentially complex and intimidating limit problems into solvable ones by reducing radicals and providing clear paths to apply additional limit-solving techniques.