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Chapter 1: Problem 20

Find the limits. \(f(x)=2 x^{2}-3 x+1, g(x)=\sqrt[3]{x+6}\) (a) \(\lim _{x \rightarrow 4} f(x)\) (b) \(\lim _{x \rightarrow 21} g(x)\) (c) \(\lim _{x \rightarrow 4} g(f(x))\)

Short Answer

Expert verified
Hence, the limits are: \n(a) \( \lim _{x \rightarrow 4} f(x) = 16\), \n(b) \( \lim _{x \rightarrow 21} g(x) = 3\), and \n(c) \( \lim _{x \rightarrow 4} g(f(x)) = \sqrt[3]{22}\)

Step by step solution

01

Find the limit of function f(x) when x approaches 4

This function is a simple polynomial function and it is continuous everywhere in its domain. Hence, to find \(\lim _{x \rightarrow 4} f(x)\), substitute \(x = 4\) directly into \(f(x)\). Thus, the limit will be \(f(4)=2(4)^2-3(4)+1=16\)
02

Find the limit of function g(x) when x approaches 21

This function is the cube root function and is also continuous everywhere in its domain. Hence, to find \(\lim _{x \rightarrow 21} g(x)\), substitute \(x = 21\) directly into \(g(x)\). So, \(\lim _{x \rightarrow 21} g(x) = g(21) = \sqrt[3]{21+6}=\sqrt[3]{27}=3\)
03

Find the limit of the composite function g(f(x)) when x approaches 4

Finding the limit of a composite function involves substituting the limit into the inner function first, and then substituting the result into the outer function. So, we first find \(f(4)\), which we have already calculated in Step 1 as \(f(4)=16\). Then, we substitute this into the function \(g(x)\), giving us \(g(f(4)) = g(16) = \sqrt[3]{16+6}=\sqrt[3]{22}\). Hence, \(\lim _{x \rightarrow 4} g(f(x)) = g(f(4))=\sqrt[3]{22}\).

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