MTematik18888

[Yusuf_Uysal2 (Yusuf Uysal)]

[sorumatikbot (Detaylı uzman cevap)]

What is the limit of the given expression as ( x ) approaches 2?

Answer:

To solve the given limit problem, we have:

\lim_{x \to 2} \frac{(\sqrt[3]{x+6} + 1)(2 - \sqrt{2x})}{x^2 - 4}

Step 1: Simplify the Denominator

The denominator ( x^2 - 4 ) can be factored as a difference of squares:

x^2 - 4 = (x - 2)(x + 2)

Step 2: Examine the Numerator

The numerator is ( (\sqrt[3]{x+6} + 1)(2 - \sqrt{2x}) ). To better understand how to simplify it or apply limits, let’s examine both parts separately:

• As ( x \to 2 ), ( \sqrt[3]{x+6} \to \sqrt[3]{8} = 2 ).
• As ( x \to 2 ), ( \sqrt{2x} \to \sqrt{4} = 2 ).

Both these substitutions result initially in:

• ( \sqrt[3]{x+6} + 1 \to 3 )
• ( 2 - \sqrt{2x} \to 0 )

Therefore, the numerator tends to ( 3 \cdot 0 = 0 ).

Step 3: Factor and Simplify Factors

The numerator, resulting in an initial ( 0 \cdot 3 = 0 ), suggests potential factorization or cancellation with the denominator is needed.

Now, let’s substitute ( x = 2 + h ) where ( h \to 0 ):

• ( \sqrt[3]{2+h+6} = \sqrt[3]{8 + h} \approx 2 + \frac{1}{12}h ) (using binomial expansion)
• ( \sqrt{2(2+h)} = \sqrt{4 + 2h} \approx 2 + \frac{1}{2}h)

Substitute these into the expansion:

(\sqrt[3]{x+6} + 1)(2-\sqrt{2x}) \approx \left(3 + \frac{1}{12}h\right)\left(-\frac{1}{2}h\right)

Resulting in:

= \left(3 + \frac{1}{12}h\right)\left(-\frac{1}{2}h\right) = -\frac{3}{2}h - \frac{1}{24}h^2

Step 4: Evaluate the Limit

Now divide this expanded numerator part by the factorized denominator ((x-2)(x+2) = h(4 + h)).

Thus,

\lim_{h \to 0} \frac{-\frac{3}{2}h - \frac{1}{24}h^2}{h(4+h)} = \lim_{h \to 0} \frac{h(-\frac{3}{2} - \frac{1}{24}h)}{h(4 + h)}

Now cancel (h):

= \lim_{h \to 0} \frac{-\frac{3}{2} - \frac{1}{24}h}{4 + h} \approx \frac{-\frac{3}{2}}{4} = -\frac{3}{8}

Final Answer:

Hence, the limit is

$$-\frac{3}{8}$$