Limits as x approaches infinity

[TommG]

Went to a tutor for help with this problem and he was stumped

need to find the limit

2 - t + sin t/ t + cos t

book says answer is -1

My attempts

I tried two things
1. I just substituted 0

2-0 + sin(0)/ 0 + cos(0)
2/1
2 wrong answer

2. I divided everything by the highest t
(2/t - t/t + sin(t)/t)/(t/t+cos(t)/t)
(2/t -1 +1)/(1 + cos(t)/t)
2t/1+cos(t)/t I got stuck right here

 

[HallsofIvy]

You titled this "Limits as x goes to infinity". But there is no "x" in what you have here. Did you mean "as t goes to infinity"?

Went to a tutor for help with this problem and he was stumped

need to find the limit

2 - t + sin t/ t + cos t

You mean (2- t+ sin t)/(t+ cos t)
\frac{2- t+ sin(t)}{t+ cos(t)}
right?
Divide both numerator and denominator by t:
\frac{\frac{2}{t}- 1+ \frac{sin(t)}{t}}{1+ \frac{cos(t)}{t}}
Now, it should be clear that, as t goes to infinity, 2/t goes to 0. Also, because for any t, sin(t) and cos(t) lie between -1 and 1, as t goes to infinity, sin(t)/t and cos(t)/t also go to 0.

book says answer is -1

My attempts

I tried two things
1. I just substituted 0

Why "0"?? Is the limit as t goes to 0 or infinity?

2-0 + sin(0)/ 0 + cos(0)
2/1
2 wrong answer

2. I divided everything by the highest t
(2/t - t/t + sin(t)/t)/(t/t+cos(t)/t)
(2/t -1 +1)/(1 + cos(t)/t)

How did "sin(t)/t)" become "1"?

2t/1+cos(t)/t I got stuck right here

 

[micromass]

Went to a tutor for help with this problem and he was stumped

need to find the limit

2 - t + sin t/ t + cos t

Can you please please please please please please use brackets! Don't write what you wrote above, but write (2 - t + sin(t))/(t + cos(t)).

 

[TommG]

You titled this "Limits as x goes to infinity". But there is no "x" in what you have here. Did you mean "as t goes to infinity"?


You mean (2- t+ sin t)/(t+ cos t)
\frac{2- t+ sin(t)}{t+ cos(t)}
right?
Divide both numerator and denominator by t:
\frac{\frac{2}{t}- 1+ \frac{sin(t)}{t}}{1+ \frac{cos(t)}{t}}
Now, it should be clear that, as t goes to infinity, 2/t goes to 0. Also, because for any t, sin(t) and cos(t) lie between -1 and 1, as t goes to infinity, sin(t)/t and cos(t)/t also go to 0.


Why "0"?? Is the limit as t goes to 0 or infinity?


How did "sin(t)/t)" become "1"?

Yes I did mean t

I used 0 because an example in the book just substituted 0

sin(t)/t became 1 because of the rule sinθ/θ = 1

 

[micromass]

sin(t)/t became 1 because of the rule sinθ/θ = 1

I don't know where you saw that rule, but it's definitely wrong.

What you probably meant was

\lim_{\theta\rightarrow 0}\frac{\sin(\theta)}{\theta} = 1

This is true but unhelpful here, since the above is the limit as $x$ approaches $0$. You want the limit as $x$ approaches infinity.

So you need to find

\lim_{\theta\rightarrow +\infty}\frac{\sin(\theta)}{\theta}

Do you know the squeeze theorem?

 

[TommG]

I don't know where you saw that rule, but it's definitely wrong.

What you probably meant was

\lim_{\theta\rightarrow 0}\frac{\sin(\theta)}{\theta} = 1

This is true but unhelpful here, since the above is the limit as $x$ approaches $0$. You want the limit as $x$ approaches infinity.

So you need to find

\lim_{\theta\rightarrow +\infty}\frac{\sin(\theta)}{\theta}

Do you know the squeeze theorem?

Yeah that was the rule I was talking about

I know the squeeze theorem but don't fully understand it.

 

[TommG]

So I substitute infinity

[(2/∞)-1 + sin∞/∞]/[1+(cos∞/∞)]
(0-1)/1
-1/1
-1

 

[Ray Vickson]

Went to a tutor for help with this problem and he was stumped

need to find the limit

2 - t + sin t/ t + cos t

book says answer is -1

My attempts

I tried two things
1. I just substituted 0

2-0 + sin(0)/ 0 + cos(0)
2/1
2 wrong answer

2. I divided everything by the highest t
(2/t - t/t + sin(t)/t)/(t/t+cos(t)/t)
(2/t -1 +1)/(1 + cos(t)/t)
2t/1+cos(t)/t I got stuck right here

You wrote your function $f(t)$ as
f(t) = 2 - t + \frac{\sin(t)}{t} + \cos(t),
whose "limit" as $t \to +\infty$ is $-\infty$, while the limit as $t \to 0$ is 3. On the other hand, if you meant
f(t) = \frac{2 - t + \sin(t)}{t + \cos(t)}
the limit as $t \to +\infty$ is -1 while the limit as $t \to 0$ is 2.
So, what is the problem, exactly? I cannot figure out anything that you are doing because of your lack of relevant parentheses.

 

[TommG]

You wrote your function $f(t)$ as
f(t) = 2 - t + \frac{\sin(t)}{t} + \cos(t),
whose "limit" as $t \to +\infty$ is $-\infty$, while the limit as $t \to 0$ is 3. On the other hand, if you meant
f(t) = \frac{2 - t + \sin(t)}{t + \cos(t)}
the limit as $t \to +\infty$ is -1 while the limit as $t \to 0$ is 2.
So, what is the problem, exactly? I cannot figure out anything that you are doing because of your lack of relevant parentheses.

yeah sorry about that ray.

I don't need help anymore thanks hallsofivy and micromass you two helped a lot.