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

[charliemagne]

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

 

[Dick]

Try substituting u=y-90 degrees.

 

[charliemagne]

Try substituting u=y-90 degrees.

what "u"?

where should I substitute it?

thank you

 

[jgens]

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

 

[charliemagne]

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?

 

[nickmai123]

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.

 

[nickmai123]

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.

 

[Dick]

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?

 

[VietDao29]

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.

 

[Dick]

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.

 

[VietDao29]

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:

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

 

[Dick]

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.