How to find lim (pi/2 - x)/cos(x) as x -> pi/2?

  • Thread starter gillgill
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  • #1
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How do i start this problem?

lim (π/2)-x all over cos x
x->π/2
 

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  • #2
cepheid
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Are you familiar with l'Hopital's rule? It seems to me that would work here.
 
  • #3
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L'Hospital's rule will work.

EDIT: cepheid beat me to it. : )
 
  • #4
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i don't know L'Hospital's rule
 
  • #5
cepheid
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Okay, well it's been a while since I learned it, so I have no idea how you would prove it is true (probably apply the def'n of a derivative). But I can tell you what it states. It usually applies to limits of one function over another: f(x)/g(x). In this case, both functions evaluate to zero at the point where you're evaluating the limit, so you get an indeterminate form: 0/0. L'Hospital's rule is especially useful in these cases:

lim f(x)/g(x) = lim f'(x)/g'(x)
x-->a ...........x -->a

The limit of the ratio of the two functions is equal to the limit of the ratio of the derivatives of the two functions. To give you some intuitive sense of why this might be so, consider another example:

lim x^2/e^x
x---> infty

This gives you the indeterminate form infty/infty

Both functions increase without bound as x becomes large. But the limit depends on which one (numerator or denominator) dominates , which depends on how fast each function grows as x increases. I.e. it depends on the derivatives. We know that the exponential function increases far more quickly than a simple power function, and L'Hospital's rule reflects that, for you can differentiate each function as many times as you want (invoke L'Hospital's each time), and you'll end up with some constant C/e^x, which tends to zero as x ---> infty. The denominator dominates, because it increases so much faster that the numerator is negligible by comparison. I hope this helps.

I'm wondering if there isn't some other way to solve it, considering you haven't been taught L'Hospital's, but were still given the problem.
 
  • #6
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how do u apply this rule to this question?
 
  • #7
Ouabache
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I also would solve using L'Hopital's rule. But if that rule had not been taught yet, another method you could use is make a substitution:

[tex] \cos x = \sqrt{1-\sin^2 x} [/tex]. Then substitute values of [tex] x [/tex] in the neighborhood of [tex] \frac{\pi}{2} [/tex], (just little bigger or smaller than) and evaluate directly.
 
  • #8
cepheid
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gillgill said:
how do u apply this rule to this question?


Well, what is f(x) (the function in the numerator) in your example? What is g(x)? Once you know, it is straightforward. Apply the formula for L'Hosptal's rule.
 
  • #9
Ouabache
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numerical solution

Gill, since you mention that you have'nt covered derivatives yet and also that you've had some problem viewing LaTeX (personal commun. from Gill), I illustrate my method for you below in plain text.

I didn't use L'Hopital's rule, just made a trig substitution then solved by evaluating x in the neighborhood of pi/2.
pi/2 = 1.57079633, for x you could plug in a value e.g. 1.56 or 1.57) and evaluate numerically. Remember to replace cos x using the identity cos x = sqrt(1-(sin x)^2).

lim (x--> pi/2) [((pi/2)-x)/cos x]
lim (x--> pi/2) [(1.57079633-1.57)/sqrt(1-sin(1.57)^2)]

do you get the idea? the solution approaches --> 1
 
  • #10
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o..icic...thx Ouabache and others...^^
 
  • #11
Galileo
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gillgill said:
How do i start this problem?

lim (π/2)-x all over cos x
x->π/2

Another way without L'Hospital is again to rewrite the equation to use the everimportant limit:

[tex]\lim_{x \to 0}\frac{\sin x}{x}=1[/tex]

Notice [itex]\sin(\pi/2-x)=-\sin(x-\pi/2)=\cos(x)[/itex]
so you can write the limit as:

[tex]\lim_{x \to \pi/2}\frac{\pi/2-x}{\sin (\pi/2-x)}[/tex]

Any substitution starting to slap you in the face?
 

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