Solving $\frac{\sin(x-1)}{x^2+x-2}$ Without L'Hopital's Rule

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    L'hopital's rule
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Discussion Overview

The discussion centers around evaluating the limit $$\lim_{x \to 1}\frac{\sin(x-1)}{x^2+x-2}$$ without using L'Hôpital's rule. Participants explore various methods of factorization and limit evaluation techniques, including the application of the limit property of sine.

Discussion Character

  • Mathematical reasoning
  • Exploratory
  • Homework-related

Main Points Raised

  • One participant initially suggests using L'Hôpital's rule but seeks alternative methods.
  • Another participant recommends factorizing the denominator and using the limit property $$\lim_{x\to 0 }\frac{\sin x}{x}=1$$.
  • There is a discussion about the correct factorization of the denominator, with a focus on identifying $(x-1)(x+2)$.
  • Participants discuss the importance of eliminating terms that cause indeterminate forms, specifically the term $(x-1)$.
  • One participant expresses confusion about how to apply the limit property after factorization, noting that the limit approaches 1 but the denominator does not approach 0.
  • Several participants confirm that after proper factorization and simplification, the limit evaluates to $$\frac{1}{3}$$.
  • There is a suggestion to use substitution, letting $$x-1=t$$, to further clarify the limit evaluation process.
  • One participant questions the validity of canceling $$\sin(x-1)$$ and $$x-1$$ in the limit expression, seeking clarification on the reasoning behind this step.
  • Another participant explains that while cancellation is permissible, it should be noted that it is based on the limit property rather than traditional fraction reduction.

Areas of Agreement / Disagreement

Participants generally agree on the final evaluation of the limit as $$\frac{1}{3}$$, but there are differing views on the methods of factorization and the appropriateness of cancellation in limit evaluation.

Contextual Notes

Some participants express uncertainty about the steps involved in factorization and limit evaluation, indicating that there may be missing assumptions or misunderstandings regarding the application of limit properties.

Petrus
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Hello,
I would like to solve this without lhopitals rule aswel( i succed get the answer 1/3 with lhopitals rule but do not go well without)
$$\lim_{x \to 1}\frac{\sin(x-1)}{x^2+x-2}$$
Any tips i would like to have
 
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Re: Trig limit

Hint , try first factorizing the denominator , and use $$\lim_{x\to 0 }\frac{\sin x}{x}=1$$.
 
Re: Trig limit

ZaidAlyafey said:
Hint , try first factorizing the denominator , and use $$\lim_{x\to 0 }\frac{\sin x}{x}=1$$.
Hello ZaidAlyafey,
I can factor out $x(x+1-2/x)$ but How can I use that..? My lim goes to 1 but to Apply that law the lim goes to 0. I se that it Will be zero on sin(1-1) but not on denominator
 
Your factorization of the denominator is not what Zaid intended. Please see:

http://www.mathhelpboards.com/f49/factoring-quadratics-3396/

for a tutorial on how to factor quadratics. You should find that the denominator factors nicely into two linear factors, one of which is $(x-1)$.

Let us know what you find. :D
 
Hello,
i keep doing bad move.. Don't get me wrong i never thought about that x-1... $(x-1)(x+2)$
Don't get me wrong i was not thinking this ohh well i learn by My misstake!
 
When solving limits of the form 0/0 , you should try to eliminate the term which makes the denominator and numerator zero , like for this example the term is (x-1) so we should foresee a way to expose it then cancel it ...
 
Petrus said:
Hello,
i keep doing bad move.. Don't get me wrong i never thought about that x-1... $(x-1)(x+2)$
Don't get me wrong i was not thinking this ohh well i learn by My misstake!

Hello Petrus,

Part of learning is making mistakes, so don't get discouraged by it. It is through our mistakes that we learn the most, as you wisely pointed out.

Can you now state the given limit the product of two limits, both of which you can evaluate?
 
MarkFL said:
Hello Petrus,

Part of learning is making mistakes, so don't get discouraged by it. It is through our mistakes that we learn the most, as you wisely pointed out.

Can you now state the given limit the product of two limits, both of which you can evaluate?
Well after factour that and simplify i get $$\lim_{x \to 1}\frac{1}{x+2}$$ = 1/3
 
Petrus said:
Well after factour that and simplify i get $$\lim_{x \to 1}\frac{1}{x+2}$$ = 1/3

You can use LH to verify your answer ...
 
  • #10
ZaidAlyafey said:
You can use LH to verify your answer ...
What does LH means?
 
  • #11
Petrus said:
What does LH means?

Oh ,sorry , I thought that was ambiguous , I meant the l'hospital's rule ...
 
  • #12
Petrus said:
Well after factour that and simplify i get $$\lim_{x \to 1}\frac{1}{x+2}$$ = 1/3

This is more what I had in mind:

$$\lim_{x\to1}\frac{\sin(x-1)}{x^2+x-2}=\lim_{x\to1} \frac{\sin(x-1)}{(x-1)(x+2)}=\lim_{x\to1} \frac{\sin(x-1)}{x-1}\cdot\lim_{x\to1}\frac{1}{x+2}=1 \cdot\frac{1}{3}=\frac{1}{3}$$

But, as long as you understand the steps involved, this is what is ultimately most important. :D
 
Last edited:
  • #13
We can as well use the substitution x-1=t then we have :

$$\lim_{t\to 0}\frac{\sin(t)}{t(t+3)}=\frac{\lim_{\,t\to 0}\frac{\sin(t)}{t}}{3}=\frac{1}{3}$$
 
  • #14
MarkFL said:
This is what more I had in mind:

$$\lim_{x\to1}\frac{\sin(x-1)}{x^2+x-2}=\lim_{x\to1} \frac{\sin(x-1)}{(x-1)(x+2)}=\lim_{x\to1} \frac{\sin(x-1)}{x-1}\cdot\lim_{x\to1}\frac{1}{x+2}=1 \cdot\frac{1}{3}=\frac{1}{3}$$

But, as long as you understand the steps involved, this is what is ultimately most important. :D
Hello Mark,
i wanted to ask if its wrong to define $$\lim_{x\to1} \frac{\sin(x-1)}{(x-1)(x+2)}=\lim_{x\to1} \frac{1}{x+2}$$ what i mean is that on paper I would draw a line on $\sin(x-1)$ and $x-1$ that means they would take out each?
 
Last edited by a moderator:
  • #15
You may casually cancel the two, as long as you are mindful that you are not dividing them out in the traditional sense of reducing a fraction, but you are only doing so in light of the result:

$$\lim_{\theta\to0}\frac{\sin(\theta)}{\theta}=1$$

On an exam, I would highly recommend making a note of this to let your professor know exactly what you are doing, otherwise you may have points deducted.
 

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