# Thread: The limit as x approaces zero of cscx - 1/x

1. ## The limit as x approaces zero of cscx - 1/x

Problem 29 from chapter 9, section 2 of the 5th edition of Calculus with Analytic Geometry by Purcell and Varberg.

My answer disagrees with the answer key, which gives $\infty$. The problem has the form $\infty - \infty$. The second and third equalities come from l'Hôpital's rule, which I applied because the second and third items have the form $\frac{0}{0}$.

$\displaystyle \lim_{x \to 0}\left(\csc x - \frac{1}{x}\right) = \lim_{x \to 0}\frac{x - \sin x}{x \sin x} = \lim_{x \to 0}\frac{1 - \cos x}{x \cos x + \sin x} = \lim_{x \to 0}\frac{\sin x}{2\cos x - x\sin x} = 0$

2. ## Re: The limit as x approaces zero of cscx - 1/x

Lookin at a graph, the limit is clearly 0.

3. ## Re: The limit as x approaches zero of cscx - 1/x

That's cheating, but both https://www.desmos.com/calculator and numpy agree. That makes three answer key errors in just this section. It had $0$ for $\displaystyle\lim_{n \to \infty}\sqrt[n]{a}$ and $\displaystyle\lim_{n \to \infty}\sqrt[n]{n}$. Still very rare overall.

4. ## Re: The limit as x approaches zero of cscx - 1/x Originally Posted by Zexuo That's cheating, but both https://www.desmos.com/calculator and numpy agree. That makes three answer key errors in just this section. It had $0$ for $\displaystyle\lim_{n \to \infty}\sqrt[n]{a}$ and $\displaystyle\lim_{n \to \infty}\sqrt[n]{n}$. Still very rare overall.
THAT is true nonsense

Look at THIS

5. ## Re: The limit as x approaces zero of cscx - 1/x

\begin{align}\lim_{x \to 0}\left(\csc{(x)} - \frac{1}{x}\right) &= \lim_{x \to 0} \frac{x - \sin{(x)}}{x \sin{(x)}} = \lim_{x \to 0} \frac{2x - \sin{(2x)}}{2x \sin{(2x)}} \\ &= \lim_{x \to 0} \left( \frac{2\cos{(x)}}{2\cos{(x)}} \cdot \frac{x - \sin{(x)}}{x \sin{(x)}} \right) \\ &= \lim_{x \to 0}\frac{2x\cos{(x)} - 2\sin{(x)}\cos{(x)}}{2x \sin{(x)}\cos{(x)}} \\ &= \lim_{x \to 0}\frac{2x\cos{(x)} - \sin{(2x)}}{x \sin{(2x)}} \\ &= 2\lim_{x \to 0}\frac{2x(\cos{(x)}-1)+2x - \sin{(2x)}}{2x \sin{(2x)}} \\ &= 2\lim_{x \to 0}\frac{2x(\cos{(x)}-1)}{2x \sin{(2x)}}+2\lim_{x \to 0}\frac{2x - \sin{(2x)}}{2x \sin{(2x)}} \\ &= \lim_{x \to 0}\frac{2x(\cos{(x)}-1)}{x \sin{(2x)}}+2\lim_{x \to 0}\frac{2x - \sin{(2x)}}{2x \sin{(2x)}} \\ &= \lim_{x \to 0} \left(\frac{2x}{\sin{(2x)}} \cdot \frac{\cos{(x)}-1}{x}\right)+2\lim_{x \to 0}\frac{2x - \sin{(2x)}}{2x \sin{(2x)}} \\ &= \cancelto{1}{\left(\lim_{x \to 0} \frac{2x}{\sin{(2x)}}\right)} \cdot \cancelto{0}{\left(\lim_{x \to 0} \frac{\cos{(x)}-1}{x}\right)}+2\cancelto{\lim_{x \to 0}\left(\csc{(x)} - \frac{1}{x}\right)}{\left(\lim_{x \to 0}\frac{2x - \sin{(2x)}}{2x \sin{(2x)}}\right)} \\ \lim_{x \to 0}\left(\csc{(x)} - \frac{1}{x}\right) &= 2\lim_{x \to 0}\left(\csc{(x)} - \frac{1}{x}\right) \\ \lim_{x \to 0}\left(\csc{(x)} - \frac{1}{x}\right) &= 0
\end{align}
Of course, the final step is only valid if the limit exists.

6. ## Re: The limit as x approaches zero of cscx - 1/x Originally Posted by Plato THAT is true nonsense

Look at THIS
Only if you are allowed a calculator to do the problem...

-Dan

7. ## Re: The limit as x approaches zero of cscx - 1/x Originally Posted by topsquark Only if you are allowed a calculator to do the problem...
Have a look at this opinion piece from the mathguy on public radio.
Then think about this question I wrote in response. Given the graph of $f(x)$ which of the following is the largest value?
$\begin{cases}a)~\int_0^6 {f(x)dx} &~b)\int_0^6 {f^2(x)dx} \\ c)~\int_0^6 {\sqrt{f(x)}dx} &d)~\int_2^6 {f^3(x)dx}\\e)~\text{cannot be determined}\end{cases}$

Over a period of years I gave this in a talk at to several meetings.
Meeting of secondary teachers of AP calculus & community college mathematics teachers.
In all of those sessions, I had only one correct answer.
I summit that this is exactly what Devlin was about in that link.
No calculator, computer , or webpage would be of any help!
Do you know the answer? Any one?

8. ## Re: The limit as x approaches zero of cscx - 1/x Originally Posted by Plato Have a look at this opinion piece from the mathguy on public radio.
Then think about this question I wrote in response. Given the graph of $f(x)$ which of the following is the largest value?
$\begin{cases}a)~\int_0^6 {f(x)dx} &~b)\int_0^6 {f^2(x)dx} \\ c)~\int_0^6 {\sqrt{f(x)}dx} &d)~\int_2^6 {f^3(x)dx}\\e)~\text{cannot be determined}\end{cases}$

Over a period of years I gave this in a talk at to several meetings.
Meeting of secondary teachers of AP calculus & community college mathematics teachers.
In all of those sessions, I had only one correct answer.
I summit that this is exactly what Devlin was about in that link.
No calculator, computer , or webpage would be of any help!
Do you know the answer? Any one?
Given that on $\displaystyle 0<y<1$, we have $\displaystyle y^a>y>y^b$ where $\displaystyle a<1<b$, I would say that c) is the correct choice here.

9. ## Re: The limit as x approaches zero of cscx - 1/x Originally Posted by MarkFL Given that on $\displaystyle 0<y<1$, we have $\displaystyle y^a>y>y^b$ where $\displaystyle a<1<b$, I would say that c) is the correct choice here.
I independently agree with you and for the same reason. Though I admit there was a Physics-y version I flunked once before so I have had some little experience with this kind of "paradox."

-Dan

10. ## Re: The limit as x approaches zero of cscx - 1/x

Sin(x) is an odd function, and it can be approximated by $\ (x + kx^3) \$
for x relatively close to 0 and for k equal to a real number.

$\displaystyle \lim_{x \to 0}\left(\csc(x) - \frac{1}{x}\right) \ =$

$\displaystyle \lim_{x \to 0}\left(\frac{1}{sin(x)} - \frac{1}{x}\right) \ \approx \$

$\displaystyle \lim_{x \to 0}\left(\frac{1}{x + kx^3} - \frac{1}{x}\right) \ = \$

$\displaystyle \lim_{x \to 0}\left(\frac{1}{x + kx^3} - \frac{1(1 + kx^2)}{x(1 + kx^2)}\right) \ = \$

$\displaystyle \lim_{x \to 0}\left(\frac{1 - 1 - kx^2}{x + kx^3} \right) \ = \$

$\displaystyle \lim_{x \to 0}\left(\frac{x(- kx)}{x(1 + kx^2)} \right) \ = \$

$\displaystyle \lim_{x \to 0}\left(\frac{- kx}{1 + kx^2} \right) \ = \$

0