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Solving a limit using only precalculus algebraic manipulations.

An elementary doubt on computing the limit $ lim limits_xto0 (sin x)^x $Calculus: L′ Hopital's RuleCalculating limits using l'Hôpital's rule.Evaluating a limit using L'Hôpital's ruleLimits of trigonometric functions: $lim _xto pi fraccos(x/2)pi ^2-x^2$Calculate $lim limits_x to 0 (tan (pi over 4-x))^cot x$ without L'Hospital's Rule or Taylor expansion.Solve this limit without using L'Hôpital's ruleLimit of $frac1xleft( 1 - fracsin xx right)$trigonometric limit using identitiesSolving limit without L'Hôpital's rule: $limlimits_x to 0 fracsqrt1+tan x-sqrt1+sin xx^3$

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1

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I'm wondering how to solve the following limit: $$limlimits_xto1fractan(x^2-1)sin(x^2-4x+3)$$ using only basic trigonometric identities from precalculus and the basic trigonometric limits $limlimits_thetato0fracsinthetatheta = 1$ and $limlimits_thetato0frac1-costhetatheta = 0$, i.e. without using L'Hôpital's rule or Taylor expansions.

I've made some headway using the substitution $u = x-1$ after some general simplifying, but cannot get a clean result.

calculus

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asked 9 hours ago

Joseph LeeJoseph Lee

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2

1

$begingroup$

I'm wondering how to solve the following limit: $$limlimits_xto1fractan(x^2-1)sin(x^2-4x+3)$$ using only basic trigonometric identities from precalculus and the basic trigonometric limits $limlimits_thetato0fracsinthetatheta = 1$ and $limlimits_thetato0frac1-costhetatheta = 0$, i.e. without using L'Hôpital's rule or Taylor expansions.

I've made some headway using the substitution $u = x-1$ after some general simplifying, but cannot get a clean result.

calculus

share|cite|improve this question

asked 9 hours ago

Joseph LeeJoseph Lee

4333 bronze badges

$endgroup$

add a comment
 | 

$begingroup$

I'm wondering how to solve the following limit: $$limlimits_xto1fractan(x^2-1)sin(x^2-4x+3)$$ using only basic trigonometric identities from precalculus and the basic trigonometric limits $limlimits_thetato0fracsinthetatheta = 1$ and $limlimits_thetato0frac1-costhetatheta = 0$, i.e. without using L'Hôpital's rule or Taylor expansions.

I've made some headway using the substitution $u = x-1$ after some general simplifying, but cannot get a clean result.

calculus

share|cite|improve this question

asked 9 hours ago

Joseph LeeJoseph Lee

4333 bronze badges

$endgroup$

share|cite|improve this question

asked 9 hours ago

Joseph LeeJoseph Lee

4333 bronze badges

share|cite|improve this question

asked 9 hours ago

Joseph LeeJoseph Lee

4333 bronze badges

asked 9 hours ago

Joseph LeeJoseph Lee

4333 bronze badges

asked 9 hours ago

Joseph LeeJoseph Lee

4333 bronze badges

2 Answers
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4

$begingroup$

We have by $y=x-1 to 0$

$$limlimits_xto1fractan(x^2-1)sin(x^2-4x+3)=limlimits_yto0fractan(y(y+2))sin(y(y-2))$$

and

$$fractan(y(y+2))sin(y(y-2))=fractan(y(y+2))y(y+2)fracy(y-2)sin(y(y-2))fracy(y+2)y(y-2)$$

share|cite|improve this answer

answered 9 hours ago

gimusigimusi

94.9k88 gold badges4646 silver badges9595 bronze badges

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5

$begingroup$

Put
$$a=x^2-1=(x-1)(x+1)$$
and
$$b=x^2-4x+3=(x-1)(x-3).$$

observe that when $ x to 1 $, $ a $ and $b $ go to zero. so

$$lim_xto 1fractan(b)bfracasin(a)fracba$$

$$=lim_xto 1frac ba=lim_xto 1fracx-3x+1=-1$$

share|cite|improve this answer

edited 7 hours ago

answered 9 hours ago

hamam_Abdallahhamam_Abdallah

39.8k22 gold badges1717 silver badges3535 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Since it is for precalculus, expliciting $ato 0$ and $bto 0$ would be great.
  $endgroup$
  – zwim
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @zwim Thanks . I did what you told me to do.
  $endgroup$
  – hamam_Abdallah
  7 hours ago
  
  

  
  
  
  

add a comment
 | 

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4

$begingroup$

We have by $y=x-1 to 0$

$$limlimits_xto1fractan(x^2-1)sin(x^2-4x+3)=limlimits_yto0fractan(y(y+2))sin(y(y-2))$$

and

$$fractan(y(y+2))sin(y(y-2))=fractan(y(y+2))y(y+2)fracy(y-2)sin(y(y-2))fracy(y+2)y(y-2)$$

share|cite|improve this answer

answered 9 hours ago

gimusigimusi

94.9k88 gold badges4646 silver badges9595 bronze badges

$endgroup$

add a comment
 | 

4

$begingroup$

We have by $y=x-1 to 0$

$$limlimits_xto1fractan(x^2-1)sin(x^2-4x+3)=limlimits_yto0fractan(y(y+2))sin(y(y-2))$$

and

$$fractan(y(y+2))sin(y(y-2))=fractan(y(y+2))y(y+2)fracy(y-2)sin(y(y-2))fracy(y+2)y(y-2)$$

share|cite|improve this answer

answered 9 hours ago

gimusigimusi

94.9k88 gold badges4646 silver badges9595 bronze badges

$endgroup$

add a comment
 | 

$begingroup$

We have by $y=x-1 to 0$

$$limlimits_xto1fractan(x^2-1)sin(x^2-4x+3)=limlimits_yto0fractan(y(y+2))sin(y(y-2))$$

and

$$fractan(y(y+2))sin(y(y-2))=fractan(y(y+2))y(y+2)fracy(y-2)sin(y(y-2))fracy(y+2)y(y-2)$$

share|cite|improve this answer

answered 9 hours ago

gimusigimusi

94.9k88 gold badges4646 silver badges9595 bronze badges

$endgroup$

share|cite|improve this answer

answered 9 hours ago

gimusigimusi

94.9k88 gold badges4646 silver badges9595 bronze badges

answered 9 hours ago

gimusigimusi

94.9k88 gold badges4646 silver badges9595 bronze badges

answered 9 hours ago

gimusigimusi

94.9k88 gold badges4646 silver badges9595 bronze badges

5

$begingroup$

Put
$$a=x^2-1=(x-1)(x+1)$$
and
$$b=x^2-4x+3=(x-1)(x-3).$$

observe that when $ x to 1 $, $ a $ and $b $ go to zero. so

$$lim_xto 1fractan(b)bfracasin(a)fracba$$

$$=lim_xto 1frac ba=lim_xto 1fracx-3x+1=-1$$

share|cite|improve this answer

edited 7 hours ago

answered 9 hours ago

hamam_Abdallahhamam_Abdallah

39.8k22 gold badges1717 silver badges3535 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Since it is for precalculus, expliciting $ato 0$ and $bto 0$ would be great.
  $endgroup$
  – zwim
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @zwim Thanks . I did what you told me to do.
  $endgroup$
  – hamam_Abdallah
  7 hours ago
  
  

  
  
  
  

add a comment
 | 

5

$begingroup$

Put
$$a=x^2-1=(x-1)(x+1)$$
and
$$b=x^2-4x+3=(x-1)(x-3).$$

observe that when $ x to 1 $, $ a $ and $b $ go to zero. so

$$lim_xto 1fractan(b)bfracasin(a)fracba$$

$$=lim_xto 1frac ba=lim_xto 1fracx-3x+1=-1$$

share|cite|improve this answer

edited 7 hours ago

answered 9 hours ago

hamam_Abdallahhamam_Abdallah

39.8k22 gold badges1717 silver badges3535 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Since it is for precalculus, expliciting $ato 0$ and $bto 0$ would be great.
  $endgroup$
  – zwim
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @zwim Thanks . I did what you told me to do.
  $endgroup$
  – hamam_Abdallah
  7 hours ago
  
  

  
  
  
  

add a comment
 | 

$begingroup$

Put
$$a=x^2-1=(x-1)(x+1)$$
and
$$b=x^2-4x+3=(x-1)(x-3).$$

observe that when $ x to 1 $, $ a $ and $b $ go to zero. so

$$lim_xto 1fractan(b)bfracasin(a)fracba$$

$$=lim_xto 1frac ba=lim_xto 1fracx-3x+1=-1$$

share|cite|improve this answer

edited 7 hours ago

answered 9 hours ago

hamam_Abdallahhamam_Abdallah

39.8k22 gold badges1717 silver badges3535 bronze badges

$endgroup$

share|cite|improve this answer

edited 7 hours ago

answered 9 hours ago

hamam_Abdallahhamam_Abdallah

39.8k22 gold badges1717 silver badges3535 bronze badges

answered 9 hours ago

hamam_Abdallahhamam_Abdallah

39.8k22 gold badges1717 silver badges3535 bronze badges

answered 9 hours ago

hamam_Abdallahhamam_Abdallah

39.8k22 gold badges1717 silver badges3535 bronze badges