Colon Blake
06/22/2023 · High School
si \( \lim _{x \rightarrow 0} f(x)=1 \) y \( \lim _{x \rightarrow 0} g(x)=-5 \), entonces \( \lim _{x \rightarrow 0} \frac{2 f(x)-g(x)}{(f(x)+7)^{\frac{1}{3}}}=\frac{7}{4} \)
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Step-by-step Solution
To solve this problem, we need to evaluate the limit of the given expression as \( x \) approaches 0. We are given that \( \lim _{x \rightarrow 0} f(x)=1 \) and \( \lim _{x \rightarrow 0} g(x)=-5 \). Let's substitute these limits into the expression and simplify.
The expression is:
\[ \lim _{x \rightarrow 0} \frac{2 f(x)-g(x)}{(f(x)+7)^{\frac{1}{3}}} \]
Substitute the limits:
\[ \lim _{x \rightarrow 0} \frac{2(1)-(-5)}{(1+7)^{\frac{1}{3}}} \]
Now, simplify the numerator and the denominator:
\[ \lim _{x \rightarrow 0} \frac{2+5}{8^{\frac{1}{3}}} \]
\[ \lim _{x \rightarrow 0} \frac{7}{8^{\frac{1}{3}}} \]
Since \( 8^{\frac{1}{3}} \) is the cube root of 8, it simplifies to 2:
\[ \lim _{x \rightarrow 0} \frac{7}{2} \]
The limit is simply the fraction:
\[ \frac{7}{2} \]
Therefore, the limit of the given expression as \( x \) approaches 0 is \( \frac{7}{2} \), not \( \frac{7}{4} \). There seems to be a mistake in the original statement of the problem.
Quick Answer
The limit is \( \frac{7}{2} \).
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