What is the indeterminate limit problem?

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Homework Help Overview

The discussion revolves around evaluating an indeterminate limit involving trigonometric functions as \( x \) approaches \( 0^+ \). The limit in question is expressed as \(\lim_{x\rightarrow 0^+} {\frac{\sqrt{2\sin^{2}(x/2)+1-\cos^2(x)}}{|x|}}\), and participants are exploring various methods to simplify or manipulate the expression without using L'Hôpital's rule.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to simplify the limit by multiplying the numerator and denominator with a related expression to eliminate the square root. They express uncertainty about their transformations and seek further insight.
  • Some participants question whether recognizing a derivative form counts as using L'Hôpital's rule and discuss alternative approaches, such as expressing the limit in terms of known limits involving sine functions.
  • Others suggest examining the expression for potential mistakes and encourage sharing more steps for clarity.

Discussion Status

The discussion is ongoing, with participants providing suggestions for alternative approaches and questioning assumptions made in the original attempts. There is no explicit consensus, but several lines of reasoning are being explored to address the limit.

Contextual Notes

Participants note the constraint of not using L'Hôpital's rule, which influences their approaches to the problem. There is also a focus on ensuring that any transformations or manipulations maintain the integrity of the limit evaluation.

khwarizm
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Homework Statement


i need to find the value of limit but its indeterminate and i can't use l'hospital's rule.
i multiplied nominator and denominator with \displaystyle\lim_{x\rightarrow 0^+} {\frac{\sqrt{2sin^{2}(x/2)+1+cos^2(x)}}{|x|}} to get rid of square root. After get i tried to make some transforms to simplify something but i couldn find anything right.


Homework Equations


This is the question:
<br /> \displaystyle\lim_{x\rightarrow 0^+} {\frac{\sqrt{2sin^{2}(x/2)+1-cos^2(x)}}{|x|}}<br />


The Attempt at a Solution


this is the last step that i came.
<br /> \displaystyle\lim_{x\rightarrow 0^+} {\frac{1-cos(x)+1-cos^{2}(x)}{|x|\sqrt{2sin^{2}(x/2)+1+cos^2(x)}}}<br />
 
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khwarizm said:

Homework Statement


i need to find the value of limit but its indeterminate and i can't use l'hospital's rule.
i multiplied nominator and denominator with \displaystyle\lim_{x\rightarrow 0^+} {\frac{\sqrt{2sin^{2}(x/2)+1+cos^2(x)}}{|x|}} to get rid of square root. After get i tried to make some transforms to simplify something but i couldn find anything right.


Homework Equations


This is the question:
<br /> \displaystyle\lim_{x\rightarrow 0^+} {\frac{\sqrt{2sin^{2}(x/2)+1-cos^2(x)}}{|x|}}<br />


The Attempt at a Solution


this is the last step that i came.
<br /> \displaystyle\lim_{x\rightarrow 0^+} {\frac{1-cos(x)+1-cos^{2}(x)}{|x|\sqrt{2sin^{2}(x/2)+1+cos^2(x)}}}<br />

Does recognising that if f(0) = 0 then <br /> f&#039;(0) = \lim_{x \to 0^{+}} \frac{f(x)-f(0)}{x} = \lim_{x \to 0^{+}} \frac{f(x)}{|x|}<br />count as "using l'Hopital's rule"?

I personally would use \cos^2(x/2) - \sin^2(x/2) = \cos \theta to express 2\sin^{2}(x/2)+1+\cos^2(x) as a quadratic in \cos x and then complete the square before differentiating.
 
$$\frac{\sqrt{2\sin^{2}(x/2)+1-\cos^2(x)}}{|x|}=\sqrt{\frac{1}{2}\left(\frac{\sin(x/2)}{x/2}\right)^2+\left(\frac{sin(x)}{x}\right)^2}$$
Do you know what
$$\lim_{x\rightarrow 0}\frac{\sin(x)}{x}=\sin^\prime(0)$$
is?

I would say recognizing
$$\lim_{x\rightarrow 0}\frac{\mathrm{f}(x+a)-\mathrm{f}(a)}{x}=\mathrm{f}^\prime(a)$$
does not count as "using l'Hopital's rule."
 
you have a mistake somewhere post more steps if you cannot find it yourself

you also could use
$$2\sin^2(x/2)+1-\cos^2(x)=2\sin^2(x/2)(2+\cos(x))$$
 

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