From aea9626ecbebf39e1e500441c4f246d440e7943e Mon Sep 17 00:00:00 2001 From: kb Date: Mon, 17 Jul 2023 21:09:44 +1000 Subject: [PATCH] CM1015 midterm WIP --- notes/study/CM1015/CM1015-midterm.md | 29 +++++++++++++++++++++++++++- 1 file changed, 28 insertions(+), 1 deletion(-) diff --git a/notes/study/CM1015/CM1015-midterm.md b/notes/study/CM1015/CM1015-midterm.md index c52aaaf..946b81a 100644 --- a/notes/study/CM1015/CM1015-midterm.md +++ b/notes/study/CM1015/CM1015-midterm.md @@ -1031,4 +1031,31 @@ Under the conditions stated, with a nominal launch we may expect the test vehicl ## Question 5 (d) -==TODO - limits== +### i. + +$$+\lim_{x\to\infty} \frac{|x - 1|}{x-1} = 1 +$$ +This can may be directory observed as for all $x > 1$ the expression will remain constant. + +## ii. + +$$+\lim_{x \to 4} \frac{2x^{3} - 128}{\sqrt{x}-2} +$$ +To reach this limit it may be initially noted that for $x = 4$ the expression is indeterminate in this form. To restate we may rationalise as follows + ++\begin{align*} +\lim_{x \to 4} \frac{2x^{3} - 128}{\sqrt{x}-2} +&= \lim_{x \to 4} \frac{2x^{3}-128}{\sqrt{x}-2} \cdot \frac{\sqrt{x} + 2}{\sqrt{x} + 2}\\ +\\ +&= \lim_{x \to 4} \frac{(2x^{3}-128)(\sqrt{x} + 2)}{x-4}\\ +\\ +&= \lim_{x \to 4} \frac{(x-4)(2x^{2}+8x+32)(\sqrt{x} + 2)}{x-4}\\ +\\ +&= \lim_{x \to 4} (2x^{2}+8x+32)(\sqrt{x} + 2)\\ +\\ +&= 384 +\end{align*} + -- 2.43.0