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| -rw-r--r-- | assignment3.tex | 12 |
1 files changed, 9 insertions, 3 deletions
diff --git a/assignment3.tex b/assignment3.tex index cb6165a..be6e3e8 100644 --- a/assignment3.tex +++ b/assignment3.tex @@ -41,8 +41,9 @@ Finally, we know $x=\sin\frac{\sqrt3}2$ gives $x=(2k+\frac13)\cdot\pi$ or $x=(2k+\frac23)\cdot\pi$ with $k\in\mathbb Z$. The only possible $x$ would then be $(2k+\frac13)\pi$ with $k=0$, so $\arcsin\frac{\sqrt3}2 = \frac\pi3$. \item \begin{align*} f'(x) &= \left(\arcsin\frac{2x}{1-x}\right)' \\ - &= \frac1{1-\left(\frac{2x}{1-x}\right)^2} \cdot \left(\frac{2x}{1-x}\right)' \\ - %todo + &= \frac1{\sqrt{1-\left(\frac{2x}{1-x}\right)^2}} \cdot \left(\frac{2x}{1-x}\right)' \\ + &= \frac1{\sqrt{1-\left(\frac{2x}{1-x}\right)^2}} \cdot \frac{(1-x)\cdot2 - 2x\cdot-1}{(1-x)^2} \\ + &= \frac2{(1-x)^2\sqrt{1-\left(\frac{2x}{1-x}\right)^2}}. \end{align*} \end{enumerate} @@ -138,7 +139,12 @@ The graph of $f$ intersect the $y$ axis at $x=0$; this gives the point $(0,f(0)) = (0, 2)$. - \item %todo + \item \begin{align*} + \lim\limits_{x\to\infty} \frac{(x-2)^2}{x+2} &= \lim\limits_{x\to\infty} \frac{1-\frac4x+\frac4{x^2}}{\frac1x+\frac2{x^2}} = \infty.\\ + \lim\limits_{x\to-\infty} \frac{(x-2)^2}{x+2} &= \lim\limits_{x\to-\infty} \frac{1-\frac4x+\frac4{x^2}}{\frac1x+\frac2{x^2}} = -\infty. + \end{align*} + + There exists no two-sided limit for $x\to-2$. This can be seen from the graph of $f$: on the right of $x=-2$, $f$ goes to $\infty$, while on the left, $f$ goes to $-\infty$. \item \begin{align*} f(x) &= \frac{(x-2)^2}{x+2} = \frac{x^2 - 4x + 4}{x+2}.\\ |