Limits

[Guest]

Hi I’m studying for my midterm final and I was reviewing some limit problems from my book but I have questions on a few of them. I would really apreciate any help I can get.

1. lim {[1/(2+x)] - (1/2)}/x
x-->0

I've tried to simplify and other things but I keep coming back to zero.

2. lim cosx/cotx
x--> pi/2

I turned cotx into cosx/sinx but I am not sure what to do from there.

Thanks in advance for your help

Carlos

 

[Unco]

Hi I’m studying for my midterm final and I was reviewing some limit problems from my book but I have questions on a few of them. I would really apreciate any help I can get.

1. lim {[1/(2+x)] - (1/2)}/x
x-->0

I've tried to simplify and other things but I keep coming back to zero.
Show your effort. Can you subtract fractions?

2. lim cosx/cotx
x--> pi/2

I turned cotx into cosx/sinx but I am not sure what to do from there.
cos(x) cancels...

 

[Guest]

Well OK for the first problem my work is as follows:

1/(2x+x^2) -1/2x I multiplied the denominator by x

2x/(4x^2+2x^3) - (2x+x^2)/(4x^2+2x^3) I found a common denominator

-x^2/(4x^2+2x^3) subtracted

-1/4+2x simplified

-1/4 plugged in x

I know that this answer is correct but only because I checked in the back of the book just now. It does not show me the step process however so I wonder if the way I got this answer is correct or my work basically.

For the second problem: I know that the cosx cancels but I don't know what to do after that because I have 1/sinx and sinx=0 at 0 so it would be undefined. Can you please help me with this one and tell me if my work for the previous one was correct?

Thank you

 

[Guest]

OH but the limit is not set to zero! It's set to pi/2 which is one so is the answer one? (for the 2nd problem)

 

[soroban]

Hello, Carlos!

\(\displaystyle \L\lim_{x\to\frac{\pi}{2}} \frac{\cos x}{\cot x}\)

I turned \(\displaystyle \cot x\) into \(\displaystyle \frac{\cos x}{\sin x}\), but I am not sure what to do from there. . Really?

This is Algebra I stuff . . .

. . \(\displaystyle \L\frac{\cos x}{\cot x} \;=\;\frac{\cos x}{\left(\frac{\cos x}{\sin x}\right)} \;=\;\sin x\)