# Funky limit solutions?

1. Mar 6, 2013

### matineesuxxx

funky limit solutions??

hey guys, this is my first post, and absolutely love this site! Anyways, I've been studying some calculus to get ahead of the game before university and I came across a few limit questions that I solved, but I don't really know HOW I solved them and by that, I mean, with some of the substitutions, I got some limits that seem intuitive to me, but I don't really know why they are that way.. Forgive my latex if it doesn't end up lookin all pretty, Im pretty new to it. So I was hoping anybody could take a look at these 3 questions and let me know whatsup?
1) \begin{equation*}
\end{equation*}

2) \begin{equation*}
\lim_{x\rightarrow 1}\frac{\sin(x-1)}{x^2+x-2}\\\\=\lim_{x\rightarrow1}\frac{\sin(x-1)}{(x+2)(x-1)}\\\\\text{let}\,a=x-1\qquad \therefore\,a+3=x+2\\\\ *\lim_{x\rightarrow 1}(x-1)=0\Rightarrow \lim_{a\rightarrow0}a=0\\\\ \therefore \lim_{x\rightarrow}\frac{\sin(x-1)}{(x+2)(x-1)}=\lim_{a\rightarrow0}\frac{\sin(a)}{a(a+3)}\\\\=(\lim_{a\rightarrow 0}\frac{1}{a+3}) (\lim_{a\rightarrow 0}\frac{\sin(a)}{a})\\\\=\frac{1}{3}
\end{equation*}

\begin{equation*}
\lim_{\theta \rightarrow 0}\frac{\sin\theta}{\theta+\tan \theta}\\\\\\=\lim_{\theta \rightarrow 0}\frac{\cos\theta \sin\theta}{\theta \cos \theta+\sin \theta}\\\\\\= \lim_{\theta \rightarrow 0}\bigg( \Big( \frac{\cos\theta\sin\theta}{\theta\cos\theta+\sin\theta}\Big)^{-1} \bigg)^{-1}\\\\\\=\lim_{\theta \rightarrow 0}\Big(\frac{\theta\cos\theta+\sin\theta}{\cos\theta \sin\theta}\Big)^{-1}\\\\\\=\lim_{\theta \rightarrow 0}\Big(\frac{\theta\cos \theta}{\cos \theta \sin \theta}+\frac{\sin\theta}{\cos\theta\sin\theta} \Big)^{-1}\\\\\\=\lim_{\theta \rightarrow 0}\Big(\frac{\theta}{\sin\theta}+\frac{1}{\cos\theta}\Big)^{-1}\\\\\\ *\lim_{\theta\rightarrow0}\Big(\frac{\theta}{\sin \theta}\Big)^{-1}=\lim_{\theta\rightarrow 0}\frac{\sin\theta}{\theta}=1 *\\\\\\= \Bigg( \lim_{\theta\rightarrow 0}\frac{\theta}{\sin\theta}+\lim_{\theta \rightarrow 0}\frac{1}{\cos\theta}\Bigg)^{-1}\\\\\\=\big(1+1)^{-1}=\frac{1}{2}
\end{equation*}

so sorry about the bad latex! I used that codecogs site, and I guess it is just a tad different?

Last edited: Mar 6, 2013
2. Mar 6, 2013

### Dick

The solutions all look good. Can you be a little more specific about what your question is?

3. Mar 6, 2013

### matineesuxxx

well, in my first one, with the:
\begin{equation*}
\lim_{x \rightarrow \infty} \frac{1}{x}=0 \Rightarrow \lim_{x \rightarrow \infty}a=0 = \lim_{a \rightarrow 0} a = 0
\end{equation*}

that is what I felt like I should do, but I was concerned about whether or not it was faulty logic?, (and I felt the same concern with that method again in the second problem), also I just felt like there should be a much much simpler way of solving the last one, because I tend to complicate things. sorry for not making my concerns known earlier :)

4. Mar 6, 2013

### Ray Vickson

Why would you think it could be faulty? Fact: as $x \to \infty$, $1/x \to 0$.

5. Mar 6, 2013

### matineesuxxx

Ok, well that makes me feel a lot better to get your input. I was told by a teacher that it was "unreasonable" to say that, then to go on and say,

\begin{equation*}
\lim_{x \rightarrow \infty}x\sin(\frac{1}{x})=\lim_{a \rightarrow 0}\frac{\sin(a)}{a}
\end{equation*}

I'm new to calculus, so I'm not very confident yet. So thanks a bunch!

6. Mar 7, 2013

### Ray Vickson

I think your teacher's statement is misleading: your argument is a good one, although maybe you need to state explicitly that you are replacing $1/x$ by $a.$