~kb/open-notes

aea9626ecbebf39e1e500441c4f246d440e7943e — kb 7 months ago 787067e
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
@@ 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

$$
\begin{align*}
\lim_{x \to 4} \frac{2x^{3} - 128}{\sqrt{x}-2}
&= \lim_{x \to 4} \frac{2x^{3}-128}{\sqrt{x}-2} \cdot \frac{\sqrt{x} + 2}{\sqrt{x} + 2}\\
\\
&= \lim_{x \to 4} \frac{(2x^{3}-128)(\sqrt{x} + 2)}{x-4}\\
\\
&= \lim_{x \to 4} \frac{(x-4)(2x^{2}+8x+32)(\sqrt{x} + 2)}{x-4}\\
\\
&= \lim_{x \to 4} (2x^{2}+8x+32)(\sqrt{x} + 2)\\
\\
&= 384
\end{align*}
$$