# Trig limit 3

#### Petrus

##### Well-known member
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
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
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

Staff member
Your factorization of the denominator is not what Zaid intended. Please see:

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. #### Petrus

##### Well-known member
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
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

Staff member
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
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
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
What does LH means?
Oh ,sorry , I thought that was ambiguous , I meant the L'Hopitals rule ...

#### MarkFL

Staff member
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. Last edited:

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
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
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. 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

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