## ~kb/open-notes

aea9626ecbebf39e1e500441c4f246d440e7943e — kb 7 months ago
CM1015 midterm WIP
1 files changed, 28 insertions(+), 1 deletions(-)

M notes/study/CM1015/CM1015-midterm.md
M notes/study/CM1015/CM1015-midterm.md => notes/study/CM1015/CM1015-midterm.md +28 -1
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## 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*}