Limits question (1 Viewer)

Shuuya · Jan 14, 2016

Could someone please help me evaluate this? I'm having a mind blank

Thanks!

InteGrand · Jan 14, 2016

Shuuya said:
Could someone please help me evaluate this? I'm having a mind blank

Thanks!

It's just 2/3. The quick way is to see that sin(2x) is "like" 2x and tan(3x) is "like" 3x when x is close to 0.

Shuuya · Jan 14, 2016

InteGrand said:
It's just 2/3. The quick way is to see that sin(2x) is "like" 2x and tan(3x) is "like" 3x when x is close to 0.

Ahh I see, thanks! So you don't need any other working out?

InteGrand · Jan 14, 2016

InteGrand · Jan 14, 2016

Shuuya said:
Ahh I see, thanks! So you don't need any other working out?

You should show working out haha. But with the quick method above, you can at least immediately see what the answer has to be, so you essentially know what kinds of manipulations to use.

Shuuya · Jan 14, 2016

Thanks!

InteGrand · Jan 14, 2016

$\noindent Basically, we have the following general result, which we can prove using the limit laws.$

$\noindent \textbf{Theorem.} Suppose $f(x) \sim F (x)$ and $g(x) \sim G (x)$ as $x\to a $ ($a$ can be a real number or $\pm \infty$, or even a one-sided limit) (see the note on notation at the bottom for the meaning of $\sim$). Suppose that $\lim _{x\to a}\frac{F(x)}{G(x)}$ exists and equals some real number $L$. Then $\lim_{x\to a}\frac {f(x)}{g(x)}$ exists and equal $L$ (assume that no denominators are identically 0 in any open neighbourhood of $a$).$

$\noindent \textbf{Proof.}$

$\noindent Note that by assumption, we have $\lim_{x\to a} \frac{f(x)}{F(x)}=1$. Also, by assumption, we have $\lim_{x\to a} \frac{g(x)}{G(x)}=1$. This implies that $\lim_{x\to a} \frac{G(x)}{g(x)}=\frac{1}{1}=1$. This is due to the reciprocal limit law which states that if $\lim_{x \to c}\varphi (x) = M$ and $M$ is a non-zero real number, then $\lim_{x \to c}\frac{1}{\varphi (x)}=\frac{1}{M}$. Now, we have$

$\begin{align*}\lim_{x\to a}\frac {f(x)}{g(x)} &= \lim_{x\to a}\frac {f(x)}{F(x)}\cdot \frac{F(x)}{G(x)}\cdot \frac{G(x)}{g(x)} \quad (\text{manipulating the limitand}) \\ &= \lim_{x\to a}\frac {f(x)}{F(x)}\cdot \lim_{x\to a} \frac{F(x)}{G(x)}\cdot \lim_{x\to a} \frac{G(x)}{g(x)} \quad (\text{using the product law of limits, which we can use here as all the limits here exist by assumption}) \\ &= 1 \cdot L \cdot 1 \quad (\text{by our assumptions}) \\ &= L,\end{align*}$

$\noindent as claimed. $\blacksquare$$

$\hrulefill$

$\noindent A note on notation: The meaning of $\phi (x)\sim \Phi (x)$ as $x\to a$ is that $\lim_{x\to a}\frac{\phi(x)}{\Phi(x)}=1$. This would read something like ``$\phi(x)$ is \textit{asymptotically equivalent to} $\Phi (x)$ as $x\to a$''. Note that if $\phi (x)\sim \Phi (x)$ as $x\to a$, then $\Phi (x) \sim \phi (x)$ as $x\to a$, due to the reciprocal limit law. In fact, $\sim$ is an equivalence relation on the set of functions of $x$, which means that in addition to this last property (which is called \textit{symmetricity}), the following are also true: as $x\to a$, if $\phi_1(x) \sim \phi_2(x)$ and $\phi_2 (x) \sim \phi_3 (x)$, then $\phi_1 (x) \sim \phi_3 (x)$ (this property is called \textit{transitivity}) and $\phi(x) \sim \phi(x)$ for any non-zero function $\phi$ (this property is called \textit{reflexivity}).$

InteGrand · Jan 14, 2016

Shuuya · Jan 15, 2016

Thanks for that super thorough explanation

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Limits question (1 Viewer)

Shuuya

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