Find the limit of (2y - 180°)/cos y as y-> 90°. How to find the answer

  • Thread starter Thread starter charliemagne
  • Start date Start date
  • Tags Tags
    Limit
charliemagne
Messages
12
Reaction score
0
please give a hint on how to solve this.

the limit of (2y - 180 degrees)/(cos y) 'as y approaches 90 degrees.'

the answer here is -2.

I need a 'hint' for me to arrive at -2.

thank you
 
Last edited:
Physics news on Phys.org


Try substituting u=y-90 degrees.
 


Dick said:
Try substituting u=y-90 degrees.

what "u"?

where should I substitute it?

thank you
 


Here, let y = u + 90 and substitute that in whenever you have a y in your equation.
 


jgens said:
Here, let y = u + 90 and substitute that in whenever you have a y in your equation.

thank you

but why should I let y be equal to u+90?

since y approaches 90 degrees, to evaluate the limit, what I have learned is to just substitute y= 90 degrees.

so how is that?
 


charliemagne said:
please give a hint on how to solve this.

the limit of (2y - 180 degrees)/(cos y) 'as y approaches 90 degrees.'

the answer here is -2.

I need a 'hint' for me to arrive at -2.

thank you

Do you know how to differentiate functions? If so, you could use L'Hôpital's rule here.
 


charliemagne said:
thank you

but why should I let y be equal to u+90?

since y approaches 90 degrees, to evaluate the limit, what I have learned is to just substitute y= 90 degrees.

so how is that?

On a different note, if you substitute in y=u+90 and u=y-90, you get rid of dividing by 0. You'll have to find the "u equivalent" of 90 because all of your variables will be with respect to u, so approaching the y value just wouldn't make sense.

EDIT: It took me a while to figure out what Dick and Jgens meant with the substitutions. I've forgotten how to take weird limits like this without using l'Hopital's rule, lol.
 


The point of the substitution is to turn part of the limit into something you know. Like sin(u)/u. cos(u+90 degrees) is -sin(u), isn't it?
 


charliemagne said:
please give a hint on how to solve this.

the limit of (2y - 180 degrees)/(cos y) 'as y approaches 90 degrees.'

the answer here is -2.

I need a 'hint' for me to arrive at -2.

thank you

Maybe I'm missing something here, but I don't really get it. Are you sure it's not:

\lim_{y \rightarrow \frac{\pi}{2}} \left( \frac{2y - \pi}{\cos y} \right)?

The well-known limit:

\lim_{x \rightarrow 0} \frac{\sin x}{x} = 1 is only true when x is in radians. And this is one of the reasons, that we use radians to do calculus work.
 
  • #10


VietDao29 said:
Maybe I'm missing something here, but I don't really get it. Are you sure it's not:

\lim_{y \rightarrow \frac{\pi}{2}} \left( \frac{2y - \pi}{\cos y} \right)?

The well-known limit:

\lim_{x \rightarrow 0} \frac{\sin x}{x} = 1 is only true when x is in radians. And this is one of the reasons, that we use radians to do calculus work.

Now that's not true. lim x->0 of sin(kx)/(kx)=1. That means lim sin(x)/x=1 is true in any units. The reason for working in radians is so that d/dx(sin(x))=cos(x). And not cos(x) times some funny unit thing.
 
  • #11


Dick said:
Now that's not true. lim x->0 of sin(kx)/(kx)=1. That means lim sin(x)/x=1 is true in any units. The reason for working in radians is so that d/dx(sin(x))=cos(x). And not cos(x) times some funny unit thing.

No, this is not true.

You can just try to plug some x near 0, in Degree Mode, and you can see the different.

In fact, the reason for d/dx(sin(x)) (x in degrees) does not equal cos(x) is because of this limit. This limit is different when x is in different modes.

Since we have:
\lim_{x \rightarrow 0} \frac{\sin x}{x} = 1 where x is in radians.

So, to find the limit:
\lim_{x \rightarrow 0} \frac{\sin x}{x}, where x is in degrees, we have to change it back to radians. The difference when changing from degrees to radians only lies in the numerator, since \sin(x_0 \mbox{ [in Rad]} ) \neq \sin(x_0 \mbox{ [in Deg]})

So:

\lim_{x \rightarrow 0} \frac{\sin x}{x} = \lim_{x \rightarrow 0} \frac{\sin \frac{\pi}{180} x}{x}
= \lim_{x \rightarrow 0} \frac{\sin \frac{\pi}{180} x}{\frac{\pi}{180}x} \times \frac{\pi}{180} = \frac{\pi}{180}.
 
  • #12


VietDao29 said:
No, this is not true.

You can just try to plug some x near 0, in Degree Mode, and you can see the different.

In fact, the reason for d/dx(sin(x)) (x in degrees) does not equal cos(x) is because of this limit. This limit is different when x is in different modes.

Since we have:
\lim_{x \rightarrow 0} \frac{\sin x}{x} = 1 where x is in radians.

So, to find the limit:
\lim_{x \rightarrow 0} \frac{\sin x}{x}, where x is in degrees, we have to change it back to radians. The difference when changing from degrees to radians only lies in the numerator, since \sin(x_0 \mbox{ [in Rad]} ) \neq \sin(x_0 \mbox{ [in Deg]})

So:

\lim_{x \rightarrow 0} \frac{\sin x}{x} = \lim_{x \rightarrow 0} \frac{\sin \frac{\pi}{180} x}{x}
= \lim_{x \rightarrow 0} \frac{\sin \frac{\pi}{180} x}{\frac{\pi}{180}x} \times \frac{\pi}{180} = \frac{\pi}{180}.

You are absolutely right. Sorry, you only convert the numerator sin(x) to radians, not the denominator.
 
Back
Top