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Math Help - Left/right hand limits.

  1. #1
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    Left/right hand limits.



    For this one, I already know that csc(x) is 1/sin(x) since csc is the inverse of sin. The limit of csc(x) or 1/sin(x) as theta approaches 0 from the right hand side is infinity. Adding a constant, which in this case is 5, should still leave the answer as infinity. Thus would infinity be the correct answer or is there more to it?



    For this one I am told to find 3 different limits for that function.

    a) As x approaches -10 from the left.
    b) As x approaches -10 from the right.
    c) As x approaches -10 from both sides.

    For (a), -10 is being approached from the left, which means that we are looking at values smaller than -10, but close to it. For example, -10.1 is a candidate. When I plug in -10.1 into the formula, I get -2231.1. When I put in -10.01, I get an even larger negative number. -10.001 yields an even LARGER negative number, and so on. Thus would it be safe to say that as x approaches -10 from the left, the limit is negative infinity?

    For (b), -10 is being approached from the right, which means that we are looking at values that are LARGER than -10, but close to it. I used -9.9 for the first candidate. When I plug it into the function, I get 2169.1. When I plug in -9.99 I get 21969.01. The larger I make it, as long as its larger than -10, the larger the answer it yields. Thus would it be correct to say that as x approaches -10 from the right, the limit is positive infinity?

    For (c), based on what I wrote above, the limit should not exist since the right hand and left hand limits differ. Plus, trying to plug in -10 itself into the function would result in the denominator being 0, thus making it undefined.

    Is my reasoning correct? If not please nudge me in the right direction. Thanks in advance to whoever helps out.
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  2. #2
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    The first one is correct, but be careful with "inverse" - formally, arcsin(x) is the inverse of sin(x), not \frac{1}{sin(x)}.

    The second one is also correct but your reasoning is faulty - checking a few values does not show what the limit is.

    To formally prove each limit, show that the numerator of the function does not tend to 0 as x tends to infinity. Then, since the denominator does tend to 0, the function will tend to positive or negative infinity, depending on which side the limit is taken from (a few details are left out for you to complete)
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  3. #3
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    Thanks for pointing that out in the first problem, I totally forgot that.

    In regards to the second one: So, for example, if I plug in -10.1 and -9.9 into the the numerator and see that I have values larger than 0 (both yielded values in the 200's), then after I check the numerator, THEN I'm clear to check the denominator (which yielded -.1 and .1 respectively)?

    Plugging in -10.1 gave me 223.11/-0.1 (which tends to negative infinity) and plugging in -9.9 gave me 216.91/0.1 (which tends to positive infinity).
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  4. #4
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    No. What I meant was that plugging in values is a faulty way to determine a limit.

    I'll show you how to do the first one, you do the rest:

    Since the numerator is continuous and \displaystyle \lim_{x \to -10} x^2 -11x +10 exists, you have that \displaystyle \lim_{x \to -10^-} x^2 -11x +10 = \lim_{x \to -10^+} x^2 -11x +10 = (-10)^2 +110 +10 = 220

    Now, note that \displaystyle \lim_{x \to -10^-} \frac{1}{x+10} = - \infty and thus \displaystyle \lim_{x \to -10^-} \frac{x^2 -11x+10}{x+10} = -\infty

    You do b).
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  5. #5
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    For the bottom line of what you did, where did you get the one from in the numerator? Or did you just put it there to represent the fact that the numerator was greater than 0?

    So for (b), wouldn't I say that since x is any value greater than -10, and when added to 10 leads to a positive decimal value? And now we have a very small positive decimal value in the denominator, and 1 [which is the numerator] divided by a very small positive decimal value leads to positive infinity?

    EDIT: So when you said plugging in values is a faulty way to do it, didn't you just mean its wrong to do it from the beginning? From what you said, we first should confirm the numerator to be a value greater than 0. After that, wouldn't we have to plug in some potential values of x into the denominator to see if the very small decimal number is either positive or negative?
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  6. #6
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    You have f(x) = \frac{x^2 -11x+10}{x+10} = \frac{1}{x+10} \cdot (x^2 -11x+10)
    You should know that if you have two functions g(x),h(x) such that g(x) \to k \ne 0 and h(x) \to \infty as x \to c, then g(x)h(x) \to \pm \infty where the sign depends on the sign of k. This is a simple application of that.

    So for (b), wouldn't I say that since x is any value greater than -10, and when added to 10 leads to a positive decimal value? And now we have a very small positive decimal value in the denominator, and 1 [which is the numerator] divided by a very small positive decimal value leads to positive infinity?
    That is correct, but it is not formal. If you're asked to prove that the limit is \infty then it will not be sufficient to say this.

    EDIT: So when you said plugging in values is a faulty way to do it, didn't you just mean its wrong to do it from the beginning? From what you said, we first should confirm the numerator to be a value greater than 0. After that, wouldn't we have to plug in some potential values of x into the denominator to see if the very small decimal number is either positive or negative?
    You can plug values to see what sign the function will have, but it is not neccesary. You can see that if, for example, x \to 10^+ then you can write  x = -10 + h where h>0 and take the limit as h \to 0, which will give \displaystyle \lim_{x \to 10^-} \frac{1}{x+10} = \lim_{h \to 0^+} \frac{1}{h} = \infty
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