Finding the limit of a very complicated trig function + some extra questions

In summary, in order to find the limit of lim (as x approaches infinite) sin[ arctan(((x^3)+2)/((x^2)+5)) + (((cosx)^2)/((|x|)^(1/2)))*arctan(2x)], you need to evaluate the limits of the composite functions inside the sine function and then substitute them back into the original limit. The arctan terms will both converge to π/2, and the cos2(x) term will approach a value between 0 and 1. Combining all these pieces together, you can see that the limit of the sine function will approach a value between -1 and 1, which agrees with the graph
  • #1
skyturnred
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Homework Statement



I am trying to find the limit of the following question:

lim (as x approaches infinite) sin[ arctan(((x^3)+2)/((x^2)+5)) + (((cosx)^2)/((|x|)^(1/2)))*arctan(2x)]

(to see the above function better, without a million brackets, please use the following link)

http://www.wolframalpha.com/input/?i=sin[+arctan%28%28%28x^3%29%2B2%29%2F%28%28x^2%29%2B5%29%29+%2B+%28%28%28cosx%29^2%29%2F%28%28|x|%29^%281%2F2%29%29%29*arctan%282x%29]


Homework Equations





The Attempt at a Solution



My attempt so far is practically non-existent! This question is part of the harder set of exercises available from my university's department for first year calc, but I figure that if I can get to a point where I can solve these kinds of problems, I should have no problem on the test!

OK, so before I even start describing my attempt, I have a quick question. Say we were given something like this:

Find lim as x approaches infinity of sin(cos(f(x))), where f(x) is of the form 0/0 or infinity over infinity, is it correct to simply isolate f(x), use hoptals rule until you come to a conclusion, and then plug it back into the limit?

Also, since the orignial question is the limit as x approaches POSITIVE infinity, can we just conclude that the |x|=x?

OK so back to the attempt, if we truly are aloud to use Hopitals rule inside of trig functions, then I get the first arctan(f(x)) to be equal to pi/2. Is this correct so far?

As for the second term within the sin, I cannot solve this! I tried making |x|=x and combined it into a single fraction. But I seem to get 0/infinity and I cannot find out how to simplify this so that it is of a form that I can use with Hopital's rule.

I realize that this is a very loaded question, but thank you so much for your help!
 
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  • #2
You are on the right track by looking at the pieces from the inside out. Yes, the arctan terms go to π/2 because their arguments go to +∞. Looking at that last term, you have the arctan going to π/2, the cos2(x) term between 0 and 1, and a √(x) in the denominator. So what does that part approach? Put it all together and what does the sine approach? Does it agree with Wolfram's picture?

Note that by taking the limits inside the functions, you are using the fact that all those functions are continuous.
 
  • #3
I strongly suggest you try and use the forums [tex] tags to typeset anything that looks that nasty, because it is really confusing to try and parse it. It'll be a worthwhile skill to have in the long run and you'll get more responses. LaTeX help

What you need here is how to evaluate the limit of a composite function. The idea is that if the limit of f(g(x)) converges, then the limit of g(x) must converge. In particular, if (for simplicity suppose f and g are continuous) [itex]\lim_{y \to \alpha}f(y)=A[/itex] and [itex]\lim_{x \to \beta}g(x)=\alpha[/itex] then

[tex]\lim_{x \to \beta}f(g(x)=\lim_{x \to \alpha}f(y)=A[/tex]

So what you will do with this problem is evaluate the inner limits, which will, in short, allow you to say the lim Sin[~] = Sin[lim ~], where ~ is just the stuff inside. You'll be doing that a couple of times since your trig function is multiply composite. Also note that since x→∞, |x|=x.

As an example, [itex]\lim_{x \to \infty}\cos(\frac{\sin(x)}{x})=\lim_{x \to 0}\cos(x)=1[/itex] since [itex]\lim_{x \to \infty}\frac{\sin(x)}{x}=0[/itex].
 

FAQ: Finding the limit of a very complicated trig function + some extra questions

1. What is the purpose of finding the limit of a trigonometric function?

The purpose of finding the limit of a trigonometric function is to determine the behavior of the function as the input values approach a certain value. This can help us understand the overall behavior of the function and make predictions about its values at specific points.

2. How do I find the limit of a trigonometric function?

To find the limit of a trigonometric function, you can use algebraic techniques such as factoring or simplifying, or you can use trigonometric identities and properties. You may also use a graphing calculator or a table of values to estimate the limit.

3. What are the common challenges when finding the limit of a complicated trigonometric function?

One of the main challenges when finding the limit of a complicated trigonometric function is identifying the proper techniques and identities to use. Another challenge is determining if the function is continuous at the limit point, as this can affect the final value of the limit.

4. Are there any special cases to consider when finding the limit of a trigonometric function?

Yes, there are a few special cases to consider when finding the limit of a trigonometric function. For example, if the function contains a rational expression with a trigonometric function in the denominator, you may need to use the Squeeze Theorem to evaluate the limit. Additionally, functions that involve infinity or undefined values may require special techniques to find the limit.

5. How can finding the limit of a trigonometric function be useful in real-world applications?

Finding the limit of a trigonometric function can be useful in real-world applications such as physics, engineering, and economics. In these fields, trigonometric functions often represent physical quantities or relationships, and understanding their behavior at specific points can help make predictions and inform decision-making.

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