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Given that \[ \lim _{x \rightarrow 2} f(x)=4 \quad \lim _{x \rightarrow 2} g(x...

Mar 22, 2024

Given that

\lim _{x \rightarrow 2} f(x)=4 \quad \lim _{x \rightarrow 2} g(x)=-4 \quad \lim _{x \rightarrow 2} h(x)=0,

find the limits, if they exist. (If an answer does not exist, enter DNE.) (a)

\lim _{x \rightarrow 2}[f(x)+2 g(x)]

(b)

\quad \lim _{x \rightarrow 2}[g(x)]^{3}

(c)

\lim _{x \rightarrow 2} \sqrt{f(x)}

(d)

\lim _{x \rightarrow 2} \frac{4 f(x)}{g(x)}

(e)

\lim _{x \rightarrow 2} \frac{g(x)}{h(x)}

(f)

\lim _{x \rightarrow 2} \frac{g(x) h(x)}{f(x)}

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Solution by Steps

step 2

\lim_{x \to 2} [f(x) + 2g(x)] = \lim_{x \to 2} f(x) + 2\lim_{x \to 2} g(x) = 4 + 2(-4) = 4 - 8

step 3

Simplifying the expression gives us 4 - 8 = -4

Key Concept

Limit of a Sum

Explanation

The limit of a sum is equal to the sum of the limits, provided the limits exist.

Solution by Steps

step 2

\lim_{x \to 2} [g(x)]^3 = (\lim_{x \to 2} g(x))^3 = (-4)^3

step 3

Simplifying the expression gives us (-4)^3 = -64

Key Concept

Limit of a Power

Explanation

The limit of a function raised to a power is the limit of the function raised to that power.

Solution by Steps

step 2

\lim_{x \to 2} \sqrt{f(x)} = \sqrt{\lim_{x \to 2} f(x)} = \sqrt{4}

step 3

Simplifying the expression gives us \sqrt{4} = 2

Key Concept

Limit of a Square Root

Explanation

The limit of the square root of a function is the square root of the limit of the function, provided the limit is non-negative.

Solution by Steps

step 2

\lim_{x \to 2} \frac{4f(x)}{g(x)} = \frac{4\lim_{x \to 2} f(x)}{\lim_{x \to 2} g(x)} = \frac{4(4)}{-4}

step 3

Simplifying the expression gives us \frac{16}{-4} = -4

Key Concept

Limit of a Quotient

Explanation

The limit of a quotient is the quotient of the limits, provided the limit of the denominator is not zero.

Solution by Steps

step 2

The expression \frac{g(x)}{h(x)} has the form \frac{-4}{0} as x approaches 2, which is undefined

Key Concept

Limits Involving Division by Zero

Explanation

If the limit of the denominator of a fraction is zero while the numerator's limit is non-zero, the limit does not exist (DNE).

Solution by Steps

step 2

Since \lim_{x \to 2} h(x) = 0, the limit of the product g(x)h(x) will also be 0

step 3

Therefore, \lim_{x \to 2} \frac{g(x)h(x)}{f(x)} = \frac{\lim_{x \to 2} g(x) \cdot \lim_{x \to 2} h(x)}{\lim_{x \to 2} f(x)} = \frac{-4 \cdot 0}{4} = 0

Key Concept

Limit of a Product over a Function

Explanation

The limit of a product over a function is the product of the limits over the limit of the function, provided all limits exist.