Elementary calculus 5

Calculus Level 1

f ( p ) = { p p 2 if , p = 4 p 2 p if , − 2 ≤ p < 4 p 2 + 2 p 3 , if , 4 < p ≤ 6 f(p) = \begin{cases} \frac{ \sqrt p}{p^2} \text {if}, p = 4 \\ \frac{p^2}{\sqrt p} \text{if}, -2 \leq p < 4 \\ \sqrt[3]{p^2 + 2p}, \text{if}, 4<p \leq 6 \end{cases}

Evaluate lim ⁡ p → 5 f ( p ) \displaystyle \lim_{p \to 5} f(p) .

Submit your answer to the nearest hundred thousand

This problem is part of this set


The answer is 3.27107.

Yellow Tomato by Yellow Tomato , 17, USA

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1 solution

Kay Xspre
Jan 13, 2016

As 4 < 5 ≤ 6 4 < 5 \leq 6 , then l i m p → 5 f ( p ) lim_{p\rightarrow5}f(p) for LHS and RHS will be the same, that is l i m p → 5 f ( p ) = l i m p → 5 ( p 2 + 2 p 3 ) = 35 3 ≈ 3.27106 \large lim_{p\rightarrow5}f(p) = lim_{p\rightarrow5}(\sqrt[3]{p^2+2p})=\sqrt[3]{35}\approx3.27106

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