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Trig limit 3

Petrus

Well-known member
Feb 21, 2013
739
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
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
Re: Trig limit

Hint , try first factorizing the denominator , and use \(\displaystyle \lim_{x\to 0 }\frac{\sin x}{x}=1\).
 

Petrus

Well-known member
Feb 21, 2013
739
Re: Trig limit

Hint , try first factorizing the denominator , and use \(\displaystyle \lim_{x\to 0 }\frac{\sin x}{x}=1\).
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
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
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
 

Petrus

Well-known member
Feb 21, 2013
739
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!
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
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 ...
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
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?
 

Petrus

Well-known member
Feb 21, 2013
739
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
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
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 ...
 

Petrus

Well-known member
Feb 21, 2013
739

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
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:

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

But, as long as you understand the steps involved, this is what is ultimately most important. :D
 
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ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
We can as well use the substitution x-1=t then we have :

\(\displaystyle \lim_{t\to 0}\frac{\sin(t)}{t(t+3)}=\frac{\lim_{\,t\to 0}\frac{\sin(t)}{t}}{3}=\frac{1}{3}\)
 

Petrus

Well-known member
Feb 21, 2013
739
This is what more I had in mind:

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

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 \(\displaystyle \lim_{x\to1} \frac{\sin(x-1)}{(x-1)(x+2)}=\lim_{x\to1} \frac{1}{x+2}\) 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:

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
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:

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

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.