Evaluate The Following Limit.Enter The Exact Answer.$\[ \lim_{x \rightarrow 2}\left(\frac{x-2}{\sqrt{x}-\sqrt{2}}\right) = \\]

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Introduction

When evaluating limits, we often encounter rational expressions with square roots in the denominator. In this article, we will focus on evaluating the limit of a rational expression with square roots, specifically the limit of (xβˆ’2xβˆ’2)\left(\frac{x-2}{\sqrt{x}-\sqrt{2}}\right) as xx approaches 2.

Understanding the Limit

To evaluate the limit, we need to understand what is meant by a limit. The limit of a function f(x)f(x) as xx approaches aa is denoted by lim⁑xβ†’af(x)\lim_{x \rightarrow a} f(x) and represents the value that the function approaches as xx gets arbitrarily close to aa. In this case, we want to find the limit of (xβˆ’2xβˆ’2)\left(\frac{x-2}{\sqrt{x}-\sqrt{2}}\right) as xx approaches 2.

Approaching the Limit

One way to approach this limit is to rationalize the denominator by multiplying both the numerator and denominator by the conjugate of the denominator. The conjugate of xβˆ’2\sqrt{x}-\sqrt{2} is x+2\sqrt{x}+\sqrt{2}. By multiplying both the numerator and denominator by x+2\sqrt{x}+\sqrt{2}, we can eliminate the square root in the denominator.

Rationalizing the Denominator

To rationalize the denominator, we multiply both the numerator and denominator by x+2\sqrt{x}+\sqrt{2}:

lim⁑xβ†’2(xβˆ’2xβˆ’2)=lim⁑xβ†’2((xβˆ’2)(x+2)(xβˆ’2)(x+2))\lim_{x \rightarrow 2}\left(\frac{x-2}{\sqrt{x}-\sqrt{2}}\right) = \lim_{x \rightarrow 2}\left(\frac{(x-2)(\sqrt{x}+\sqrt{2})}{(\sqrt{x}-\sqrt{2})(\sqrt{x}+\sqrt{2})}\right)

Simplifying the Expression

Now, we can simplify the expression by expanding the numerator and denominator:

lim⁑xβ†’2((xβˆ’2)(x+2)(xβˆ’2)(x+2))=lim⁑xβ†’2(xx+x2βˆ’2xβˆ’22xβˆ’2)\lim_{x \rightarrow 2}\left(\frac{(x-2)(\sqrt{x}+\sqrt{2})}{(\sqrt{x}-\sqrt{2})(\sqrt{x}+\sqrt{2})}\right) = \lim_{x \rightarrow 2}\left(\frac{x\sqrt{x}+x\sqrt{2}-2\sqrt{x}-2\sqrt{2}}{x-2}\right)

Cancelling Out Terms

We can cancel out the terms xβˆ’2x-2 in the numerator and denominator:

lim⁑xβ†’2(xx+x2βˆ’2xβˆ’22xβˆ’2)=lim⁑xβ†’2(x+2βˆ’2x+22xβˆ’2)\lim_{x \rightarrow 2}\left(\frac{x\sqrt{x}+x\sqrt{2}-2\sqrt{x}-2\sqrt{2}}{x-2}\right) = \lim_{x \rightarrow 2}\left(\sqrt{x}+\sqrt{2}-\frac{2\sqrt{x}+2\sqrt{2}}{x-2}\right)

Evaluating the Limit

Now, we can evaluate the limit by substituting x=2x=2 into the expression:

lim⁑xβ†’2(x+2βˆ’2x+22xβˆ’2)=2+2βˆ’22+222βˆ’2\lim_{x \rightarrow 2}\left(\sqrt{x}+\sqrt{2}-\frac{2\sqrt{x}+2\sqrt{2}}{x-2}\right) = \sqrt{2}+\sqrt{2}-\frac{2\sqrt{2}+2\sqrt{2}}{2-2}

Simplifying the Expression

We can simplify the expression by evaluating the fraction:

2+2βˆ’22+222βˆ’2=2+2βˆ’420\sqrt{2}+\sqrt{2}-\frac{2\sqrt{2}+2\sqrt{2}}{2-2} = \sqrt{2}+\sqrt{2}-\frac{4\sqrt{2}}{0}

Indeterminate Form

We have an indeterminate form 00\frac{0}{0}, which means that we need to use L'Hopital's rule to evaluate the limit.

L'Hopital's Rule

L'Hopital's rule states that if we have an indeterminate form 00\frac{0}{0}, we can evaluate the limit by taking the derivative of the numerator and denominator separately.

Applying L'Hopital's Rule

We can apply L'Hopital's rule to our expression:

lim⁑xβ†’2(x+2βˆ’2x+22xβˆ’2)=lim⁑xβ†’2(12xβˆ’22(xβˆ’2))\lim_{x \rightarrow 2}\left(\sqrt{x}+\sqrt{2}-\frac{2\sqrt{x}+2\sqrt{2}}{x-2}\right) = \lim_{x \rightarrow 2}\left(\frac{1}{2\sqrt{x}}-\frac{2}{2(x-2)}\right)

Simplifying the Expression

We can simplify the expression by evaluating the derivatives:

lim⁑xβ†’2(12xβˆ’22(xβˆ’2))=lim⁑xβ†’2(12xβˆ’1xβˆ’2)\lim_{x \rightarrow 2}\left(\frac{1}{2\sqrt{x}}-\frac{2}{2(x-2)}\right) = \lim_{x \rightarrow 2}\left(\frac{1}{2\sqrt{x}}-\frac{1}{x-2}\right)

Evaluating the Limit

Now, we can evaluate the limit by substituting x=2x=2 into the expression:

lim⁑xβ†’2(12xβˆ’1xβˆ’2)=122βˆ’12βˆ’2\lim_{x \rightarrow 2}\left(\frac{1}{2\sqrt{x}}-\frac{1}{x-2}\right) = \frac{1}{2\sqrt{2}}-\frac{1}{2-2}

Simplifying the Expression

We can simplify the expression by evaluating the fraction:

122βˆ’12βˆ’2=122βˆ’10\frac{1}{2\sqrt{2}}-\frac{1}{2-2} = \frac{1}{2\sqrt{2}}-\frac{1}{0}

Indeterminate Form

We have an indeterminate form 00\frac{0}{0}, which means that we need to use L'Hopital's rule to evaluate the limit.

L'Hopital's Rule

L'Hopital's rule states that if we have an indeterminate form 00\frac{0}{0}, we can evaluate the limit by taking the derivative of the numerator and denominator separately.

Applying L'Hopital's Rule

We can apply L'Hopital's rule to our expression:

lim⁑xβ†’2(12xβˆ’1xβˆ’2)=lim⁑xβ†’2(βˆ’14xxβˆ’1(xβˆ’2)2)\lim_{x \rightarrow 2}\left(\frac{1}{2\sqrt{x}}-\frac{1}{x-2}\right) = \lim_{x \rightarrow 2}\left(\frac{-1}{4x\sqrt{x}}-\frac{1}{(x-2)^2}\right)

Simplifying the Expression

We can simplify the expression by evaluating the derivatives:

lim⁑xβ†’2(βˆ’14xxβˆ’1(xβˆ’2)2)=lim⁑xβ†’2(βˆ’14xxβˆ’14(xβˆ’2)2)\lim_{x \rightarrow 2}\left(\frac{-1}{4x\sqrt{x}}-\frac{1}{(x-2)^2}\right) = \lim_{x \rightarrow 2}\left(\frac{-1}{4x\sqrt{x}}-\frac{1}{4(x-2)^2}\right)

Evaluating the Limit

Now, we can evaluate the limit by substituting x=2x=2 into the expression:

lim⁑xβ†’2(βˆ’14xxβˆ’14(xβˆ’2)2)=βˆ’14(2)2βˆ’14(2βˆ’2)2\lim_{x \rightarrow 2}\left(\frac{-1}{4x\sqrt{x}}-\frac{1}{4(x-2)^2}\right) = \frac{-1}{4(2)\sqrt{2}}-\frac{1}{4(2-2)^2}

Simplifying the Expression

We can simplify the expression by evaluating the fraction:

βˆ’14(2)2βˆ’14(2βˆ’2)2=βˆ’182βˆ’10\frac{-1}{4(2)\sqrt{2}}-\frac{1}{4(2-2)^2} = \frac{-1}{8\sqrt{2}}-\frac{1}{0}

Indeterminate Form

We have an indeterminate form 00\frac{0}{0}, which means that we need to use L'Hopital's rule to evaluate the limit.

L'Hopital's Rule

L'Hopital's rule states that if we have an indeterminate form 00\frac{0}{0}, we can evaluate the limit by taking the derivative of the numerator and denominator separately.

Applying L'Hopital's Rule

We can apply L'Hopital's rule to our expression:

lim⁑xβ†’2(βˆ’18xβˆ’14(xβˆ’2)2)=lim⁑xβ†’2(116xx+24(xβˆ’2)3)\lim_{x \rightarrow 2}\left(\frac{-1}{8\sqrt{x}}-\frac{1}{4(x-2)^2}\right) = \lim_{x \rightarrow 2}\left(\frac{1}{16x\sqrt{x}}+\frac{2}{4(x-2)^3}\right)

Simplifying the Expression

We can simplify the expression by evaluating the derivatives:

lim⁑xβ†’2(116xx+24(xβˆ’2)3)=lim⁑xβ†’2(116xx+12(xβˆ’2)3)\lim_{x \rightarrow 2}\left(\frac{1}{16x\sqrt{x}}+\frac{2}{4(x-2)^3}\right) = \lim_{x \rightarrow 2}\left(\frac{1}{16x\sqrt{x}}+\frac{1}{2(x-2)^3}\right)

Evaluating the Limit

Now, we can evaluate the limit by substituting x=2x=2 into the expression:

lim⁑xβ†’2(116xx+12(xβˆ’2)3)=116(2)2+12(2βˆ’2)3\lim_{x \rightarrow 2}\left(\frac{1}{16x\sqrt{x}}+\frac{1}{2(x-2)^3}\right) = \frac{1}{16(2)\sqrt{2}}+\frac{1}{2(2-2)^3}

Simplifying the Expression

We can simplify the expression by evaluating the fraction:

\frac{1}{16(2)\sqrt{2}}<br/> # **Evaluating the Limit of a Rational Expression with Square Roots: Q&A** ## **Introduction** In our previous article, we evaluated the limit of a rational expression with square roots, specifically the limit of $\left(\frac{x-2}{\sqrt{x}-\sqrt{2}}\right)$ as $x$ approaches 2. In this article, we will provide a Q&A section to help clarify any doubts and provide additional information on evaluating limits with rational expressions and square roots. ## **Q: What is the limit of a rational expression with square roots?** A: The limit of a rational expression with square roots is the value that the expression approaches as the input (or independent variable) gets arbitrarily close to a certain value. In this case, we want to find the limit of $\left(\frac{x-2}{\sqrt{x}-\sqrt{2}}\right)$ as $x$ approaches 2. ## **Q: How do I rationalize the denominator of a rational expression with square roots?** A: To rationalize the denominator, you can multiply both the numerator and denominator by the conjugate of the denominator. The conjugate of $\sqrt{x}-\sqrt{2}$ is $\sqrt{x}+\sqrt{2}$. By multiplying both the numerator and denominator by $\sqrt{x}+\sqrt{2}$, you can eliminate the square root in the denominator. ## **Q: What is L'Hopital's rule, and how do I apply it?** A: L'Hopital's rule is a technique used to evaluate limits of the form $\frac{0}{0}$ or $\frac{\infty}{\infty}$. To apply L'Hopital's rule, you take the derivative of the numerator and denominator separately and then evaluate the limit of the resulting expression. ## **Q: Why do I need to use L'Hopital's rule in this case?** A: In this case, we have an indeterminate form $\frac{0}{0}$, which means that we need to use L'Hopital's rule to evaluate the limit. By taking the derivative of the numerator and denominator separately, we can eliminate the indeterminate form and evaluate the limit. ## **Q: Can I use L'Hopital's rule multiple times?** A: Yes, you can use L'Hopital's rule multiple times if the limit is still indeterminate after the first application. However, be careful not to apply L'Hopital's rule too many times, as this can lead to an infinite loop. ## **Q: What are some common mistakes to avoid when evaluating limits with rational expressions and square roots?** A: Some common mistakes to avoid when evaluating limits with rational expressions and square roots include: * Not rationalizing the denominator * Not applying L'Hopital's rule when necessary * Applying L'Hopital's rule too many times * Not checking for indeterminate forms ## **Q: How do I check for indeterminate forms?** A: To check for indeterminate forms, you can plug in the value that the input is approaching into the expression. If the expression is still undefined, then you have an indeterminate form. ## **Q: Can I use other techniques to evaluate limits with rational expressions and square roots?** A: Yes, there are other techniques you can use to evaluate limits with rational expressions and square roots, such as factoring or canceling out terms. However, L'Hopital's rule is often the most effective technique to use in these cases. ## **Conclusion** Evaluating limits with rational expressions and square roots can be challenging, but with the right techniques and practice, you can master these types of problems. Remember to rationalize the denominator, apply L'Hopital's rule when necessary, and check for indeterminate forms. With these tips and techniques, you'll be well on your way to becoming a limit-evaluation master!