Question

lim as x-> 5pi/2 of (sin^2x + 6sinx + 5)/sin^2x-1

Please explain HOW you do it...

Answer 1

First I would simplify this expression and see what you can see.

Answer 2

hint: factor in terms of sin x

Answer 3

I got it to (sinx + 5)(sinx + 1) / (sinx+1)(sinx-1)

Answer 4

but cancelling doesn't help

Answer 5

right, now cancel one of these terms. And then try and take the limit.

Answer 6

because sin(5pi/2) is 1
and 1-1 is 0..which makes the denominator 0

Answer 7

yes, the denominator approaches zero from below. And the numerator is always positive.

Hence the limit is improper and what is it?

Answer 8

I don't understand

Answer 9

The numerator is positive, the denominator is negative and approaches zero as x --> 5pi/2.

Hence the limit must be ... what?

Answer 10

negative infinity? but how do you know the denominator is negative?

Answer 11

Yes, -infty.

Because for all values of x, \( \sin x \leq 1 \) and in a neighborhood around x = 5pi/2--but not including 5pi/2, sin x is strictly less than 1. Therefore sin x - 1 < 0

Answer 12

what is "\( \sin x \leq 1 \)"

Answer 13

sin x is less than or equal to 1.

Answer 14

ok and how do you know it's less than 0?

Answer 15

because sin(5pi/2) = 1 but for any x near 5pi/2, sin(x) < 1. Hence for such values of x:
sin(x) - 1 < 0

Answer 16

Here's an alternative way. sin^2 x - 1 = -cos^2 x. Now take the limit of the whole original expression and you'll find the same answer.

Answer 17

OH wait I think I understand

Answer 18

at 5pi/2, on the unit circle, sin is at its highest

Answer 19

yes

Answer 20

if you approach it from the left or right, it has to be less than 1 because it is a circle

Answer 21

ok I understand how it is negative, why is it infinity?

Answer 22

negative infinity. Because the numerator is positive, and the denominator is negative going to zero.

Answer 23

yes I mean where does the infinity part come from

Answer 24

i understand that the answer has to be negative

Answer 25

What's the limit of
\[ \lim_{x \to 0} \frac{17}{-x^2} \]

Answer 26

lim x-> 0 of 17/-x^2 ?

sorry, but for some reason your text is all messed up

Answer 27

i think it's because the equation part of openstudy isn't working

Answer 28

Yes, that limit is -infty

Answer 29

yes because the numerator is constant and the bottom is smaller and smaller but still negative

Answer 30

something's wrong with your browser then.

But in any case, you see why that's the answer? The same thing is true here with your limit.

lim x -> 5pi/2 of (sin x + 5)/(sin x - 1)

The numerator is ALWAYS positive, but the denominator is negative and going to zero.

Answer 31

is that problem supposed to be really tricky?

Answer 32

haha I understand it but i feel like it's very obscure

Answer 33

It tests your understanding. Try it the other way I proposed using
sin^2 x - 1 = -cos^2 x

You might find that a bit more intuitive.

But in any case, one of the reason we do these problems is to work out the kinks in our conceptual understanding.

Answer 34

yeah it really blew my mind when I finally understood it, thanks!