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How do i start this problem?

lim (π/2)-x all over cos x

x->π/2

lim (π/2)-x all over cos x

x->π/2

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- 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

lim (π/2)-x all over cos x

x->π/2

- #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. : )

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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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)

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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[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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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

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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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