Solving Two Limit Problems: Find \lim_{x\rightarrow 0, \frac{\pi}{4}}

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Homework Help Overview

The discussion revolves around two limit problems involving trigonometric functions. The first limit is as \( x \) approaches 0 for the expression \( \frac{\sin^{2}(\frac{x}{2})}{\sin(x)} \), and the second limit is as \( x \) approaches \( \frac{\pi}{4} \) for the expression \( \frac{1 - \tan(x)}{\sin(x) - \cos(x)} \).

Discussion Character

  • Exploratory, Assumption checking, Mixed

Approaches and Questions Raised

  • Participants discuss various approaches to solve the limits, including the use of l'Hospital's Rule and trigonometric identities. There are questions about how to handle the limits without l'Hospital's Rule, and some participants express uncertainty about the methods they should use.

Discussion Status

Some participants have suggested using l'Hospital's Rule for both problems, while others are exploring alternative methods. There is acknowledgment of the need to understand the limits without relying on l'Hospital's Rule, and some participants are actively seeking to learn more about the topic.

Contextual Notes

There is mention of course constraints regarding the use of l'Hospital's Rule, with some participants indicating that it may not be covered in their curriculum. This has led to discussions about the importance of solving the limits through other means.

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Hello,

I am having difficulties with two limit problems and hopefully someone can show me a trick or two that I don't know!

1) find [tex]\lim_{x\rightarrow 0}\frac{sin^{2}(\frac{x}{2})}{sin(x)}[/tex]

Is it just me or is this thing a beast of a limit. I just can't get the denominator to equal anything other then 0!

I tried using the half angle identity on the top but it simply doesn't help...

2)find [tex]\lim_{x\rightarrow \frac{\pi}{4}}\frac{1-tan(x)}{sin(x)-cos(x)}[/tex] This one I don't even know where to start, I know that I need to get 1-cos(x)/x or cos(x)-1(x) and/or sin(x)/x or x/sin(x)... but I don't see a way.

PS- that's x approaches x/2(not pi...) sigh (for q1) and PI/4 (for q2) btw. it's small. :(

Help! :bugeye:
 
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For (1) Use l'hospital's RuleFor (2) rewrite [tex]1-\tan x[/tex] as [tex]1 - \frac{\sin x}{\cos x}[/tex]
 
I don't know l'hospital's Rule. Is it easy to learn? If so I will just look it up and try to learn it :)

For q2, I will try it, thanks.
 
Learned it already, :) well, sorta, :) seems fairly straight forward.
 
I'd be interested in knowing how can 1) be done without l'Hospital. The most natural first step would be to try the substitution

[tex]\sin^2(\frac{x}{2})=\frac{1-\cos x}{2}[/tex]

but I don't see where to go from there...
 
Yea, it's beastly. I'll ask my teacher and post it :)
 
I used l'hospital's for both #1 and #2, :). Very straightforward and immensely useful!
 
mmmh yea but I hope you're aware that if you haven't learned l'Hospital's rule in class yet, and use it in the exam, that's 0 for you. So you'd better learn to do the limits without it.
 
In the supplementary note package that came with my course it says that l'hospitals is outside the scope of the course, but says that it can be used. I just never bothered to learn it until now. But I do agree with you 110% that I need to do them both ways if I want to use l'hospitals. Not because I will necessarily lose marks, but because the practice is needed.
I think I will do a bunch of limit problems since I still can't do #1 without L'Hospitals...
 
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