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What is the value of $lim_x to 1^+ fracsqrtx^2-1+x-1 sqrt4x-4+x^2-1$?

Finding the limit of $fracsqrtxsqrtx+sinsqrtx$Calculate $lim_x to infty x - sqrtx^2 + 2x$ without derivationsFind $lim_xto-inftyfracxsqrtx^2+2$Find $lim_nto inftysqrt[n]fracsum_i=1^p a_i^np$Find the value of : $lim_xtoinftyfracsqrtx-1 - sqrtx-2sqrtx-2 - sqrtx-3$Find the limit $lim_t to 9 frac3-sqrtt9-t$$lim_n to infty fracsqrt1 + sqrt2 + … + sqrtnnsqrtn$Find: $lim_xtoinfty fracsqrtxsqrtx+sqrtx+sqrtx.$How to solve $lim_xto1=fracx^2+x-21-sqrtx$?finding value of $lim_nrightarrow inftysqrt[n]frac(27)^n(n!)^3(3n)!$

1

$begingroup$

$$lim_x to 1^+ fracsqrtx^2-1+x-1 sqrt4x-4+x^2-1 = ?$$

I have done these steps to find the answer:

1. $x^2-1=(x+1)(x-1)$




2. $sqrt4x-4=2sqrtx-1$




3. $displaystylelim_x to 1^+ fracsqrtx^2-1+x-1 sqrt4x-4+x^2-1 = lim_x to 1^+ fracsqrt (x+1)(x-1) +x-1 2sqrtx-1 + (x+1)(x-1) $

So how do I remove what causes the hole function not to become $frac00$ and solve the limit?

limits

share|cite|improve this question

edited Mar 16 at 11:36

egreg

185k1486206

asked Mar 16 at 8:01

AquamanAquaman

133

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The numerator tends to $2$, the denominator to $0$.
  $endgroup$
  – Lord Shark the Unknown
  Mar 16 at 8:04
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Did you mean $$lim_x to 1^+ fracsqrtx^2-1+x-1 sqrt4x-4+x^2-1 ?$$
  $endgroup$
  – Lord Shark the Unknown
  Mar 16 at 8:05
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes I have edited it now!
  $endgroup$
  – Aquaman
  Mar 16 at 8:09
  

  
  
  
  

add a comment | 

1

$begingroup$

$$lim_x to 1^+ fracsqrtx^2-1+x-1 sqrt4x-4+x^2-1 = ?$$

I have done these steps to find the answer:

1. $x^2-1=(x+1)(x-1)$




2. $sqrt4x-4=2sqrtx-1$




3. $displaystylelim_x to 1^+ fracsqrtx^2-1+x-1 sqrt4x-4+x^2-1 = lim_x to 1^+ fracsqrt (x+1)(x-1) +x-1 2sqrtx-1 + (x+1)(x-1) $

So how do I remove what causes the hole function not to become $frac00$ and solve the limit?

limits

share|cite|improve this question

edited Mar 16 at 11:36

egreg

185k1486206

asked Mar 16 at 8:01

AquamanAquaman

133

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The numerator tends to $2$, the denominator to $0$.
  $endgroup$
  – Lord Shark the Unknown
  Mar 16 at 8:04
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Did you mean $$lim_x to 1^+ fracsqrtx^2-1+x-1 sqrt4x-4+x^2-1 ?$$
  $endgroup$
  – Lord Shark the Unknown
  Mar 16 at 8:05
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes I have edited it now!
  $endgroup$
  – Aquaman
  Mar 16 at 8:09
  

  
  
  
  

add a comment | 

$begingroup$

$$lim_x to 1^+ fracsqrtx^2-1+x-1 sqrt4x-4+x^2-1 = ?$$

I have done these steps to find the answer:

1. $x^2-1=(x+1)(x-1)$




2. $sqrt4x-4=2sqrtx-1$




3. $displaystylelim_x to 1^+ fracsqrtx^2-1+x-1 sqrt4x-4+x^2-1 = lim_x to 1^+ fracsqrt (x+1)(x-1) +x-1 2sqrtx-1 + (x+1)(x-1) $

So how do I remove what causes the hole function not to become $frac00$ and solve the limit?

limits

share|cite|improve this question

edited Mar 16 at 11:36

egreg

185k1486206

asked Mar 16 at 8:01

AquamanAquaman

133

$endgroup$

share|cite|improve this question

edited Mar 16 at 11:36

egreg

185k1486206

asked Mar 16 at 8:01

AquamanAquaman

133

share|cite|improve this question

edited Mar 16 at 11:36

egreg

185k1486206

asked Mar 16 at 8:01

AquamanAquaman

133

edited Mar 16 at 11:36

egreg

185k1486206

edited Mar 16 at 11:36

egreg

185k1486206

asked Mar 16 at 8:01

AquamanAquaman

133

asked Mar 16 at 8:01

AquamanAquaman

133

3 Answers
3

active

oldest

votes

2

$begingroup$

The expression is not of the form $0/0$ notice carefully the denominator goes to $0$ where the numerator is finite.

$$lim_xto 1^+dfracsqrtx^2-1+x+1sqrt4x-4+x^2-1to infty$$

This graph confirms the result: https://www.desmos.com/calculator/ejj9xiyjcf

---

After the OP's edit: $$dfracsqrtx^2-1+x-12sqrtx-1+x^2-1=dfracsqrtx-1sqrtx-1cdotdfracsqrtx+1+sqrtx-12+(x-1)^3/2(x+1) \ implies lim_xto 1^+dfracsqrtx+1+sqrtx-12+(x-1)^3/2(x+1)to dfracsqrt22$$

share|cite|improve this answer

edited Mar 17 at 6:13

answered Mar 16 at 8:04

Paras KhoslaParas Khosla

2,643323

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Are you sure? I am not!
  $endgroup$
  – Aquaman
  Mar 16 at 8:04
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I think that Zeros are limit kinds of zero! I mean they are $0^+$ & $0^-$
  $endgroup$
  – Aquaman
  Mar 16 at 8:06
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I think that we have to simplify the function then solve it!
  $endgroup$
  – Aquaman
  Mar 16 at 8:07
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  AFTER REMOVING + and replacing - it will change . Sorry ive edited the question!
  $endgroup$
  – Aquaman
  Mar 16 at 8:12
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Ok @Aquaman I've edited my answer too. Cheers:- ))
  $endgroup$
  – Paras Khosla
  Mar 16 at 8:14
  

  
  
  
  

add a comment | 

2

$begingroup$

I assume:
$$lim_x to 1^+ fracsqrtx^2-1+xcolorred-1 sqrt4x-4+x^2-1 = lim_x to 1^+ fracsqrtx-1cdot sqrtx+1+(sqrtx-1)^2 sqrt4x-4+(sqrtx-1)^2(x+1) =\
lim_x to 1^+ fracsqrtx-1cdot (sqrtx+1+sqrtx-1)sqrtx-1cdot (2+(sqrtx-1)(x+1)) =fracsqrt22.$$

share|cite|improve this answer

answered Mar 16 at 8:08

farruhotafarruhota

21.5k2842

$endgroup$

add a comment | 

0

$begingroup$

Set $x-1=t^2$, which you can because the limit is for $xto1^+$. Then the limit becomes
$$
lim_tto0^+fractsqrtt^2+2+t^22t+t^2(t^2+2)=
lim_tto0^+fracsqrtt^2+2+t2+t(t^2+2)
$$

share|cite|improve this answer

answered Mar 16 at 11:39

egregegreg

185k1486206

$endgroup$

add a comment | 

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2

$begingroup$

The expression is not of the form $0/0$ notice carefully the denominator goes to $0$ where the numerator is finite.

$$lim_xto 1^+dfracsqrtx^2-1+x+1sqrt4x-4+x^2-1to infty$$

This graph confirms the result: https://www.desmos.com/calculator/ejj9xiyjcf

---

After the OP's edit: $$dfracsqrtx^2-1+x-12sqrtx-1+x^2-1=dfracsqrtx-1sqrtx-1cdotdfracsqrtx+1+sqrtx-12+(x-1)^3/2(x+1) \ implies lim_xto 1^+dfracsqrtx+1+sqrtx-12+(x-1)^3/2(x+1)to dfracsqrt22$$

share|cite|improve this answer

edited Mar 17 at 6:13

answered Mar 16 at 8:04

Paras KhoslaParas Khosla

2,643323

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Are you sure? I am not!
  $endgroup$
  – Aquaman
  Mar 16 at 8:04
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I think that Zeros are limit kinds of zero! I mean they are $0^+$ & $0^-$
  $endgroup$
  – Aquaman
  Mar 16 at 8:06
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I think that we have to simplify the function then solve it!
  $endgroup$
  – Aquaman
  Mar 16 at 8:07
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  AFTER REMOVING + and replacing - it will change . Sorry ive edited the question!
  $endgroup$
  – Aquaman
  Mar 16 at 8:12
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Ok @Aquaman I've edited my answer too. Cheers:- ))
  $endgroup$
  – Paras Khosla
  Mar 16 at 8:14
  

  
  
  
  

add a comment | 

2

$begingroup$

The expression is not of the form $0/0$ notice carefully the denominator goes to $0$ where the numerator is finite.

$$lim_xto 1^+dfracsqrtx^2-1+x+1sqrt4x-4+x^2-1to infty$$

This graph confirms the result: https://www.desmos.com/calculator/ejj9xiyjcf

---

After the OP's edit: $$dfracsqrtx^2-1+x-12sqrtx-1+x^2-1=dfracsqrtx-1sqrtx-1cdotdfracsqrtx+1+sqrtx-12+(x-1)^3/2(x+1) \ implies lim_xto 1^+dfracsqrtx+1+sqrtx-12+(x-1)^3/2(x+1)to dfracsqrt22$$

share|cite|improve this answer

edited Mar 17 at 6:13

answered Mar 16 at 8:04

Paras KhoslaParas Khosla

2,643323

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Are you sure? I am not!
  $endgroup$
  – Aquaman
  Mar 16 at 8:04
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I think that Zeros are limit kinds of zero! I mean they are $0^+$ & $0^-$
  $endgroup$
  – Aquaman
  Mar 16 at 8:06
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I think that we have to simplify the function then solve it!
  $endgroup$
  – Aquaman
  Mar 16 at 8:07
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  AFTER REMOVING + and replacing - it will change . Sorry ive edited the question!
  $endgroup$
  – Aquaman
  Mar 16 at 8:12
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Ok @Aquaman I've edited my answer too. Cheers:- ))
  $endgroup$
  – Paras Khosla
  Mar 16 at 8:14
  

  
  
  
  

add a comment | 

$begingroup$

The expression is not of the form $0/0$ notice carefully the denominator goes to $0$ where the numerator is finite.

$$lim_xto 1^+dfracsqrtx^2-1+x+1sqrt4x-4+x^2-1to infty$$

This graph confirms the result: https://www.desmos.com/calculator/ejj9xiyjcf

---

After the OP's edit: $$dfracsqrtx^2-1+x-12sqrtx-1+x^2-1=dfracsqrtx-1sqrtx-1cdotdfracsqrtx+1+sqrtx-12+(x-1)^3/2(x+1) \ implies lim_xto 1^+dfracsqrtx+1+sqrtx-12+(x-1)^3/2(x+1)to dfracsqrt22$$

share|cite|improve this answer

edited Mar 17 at 6:13

answered Mar 16 at 8:04

Paras KhoslaParas Khosla

2,643323

$endgroup$

share|cite|improve this answer

edited Mar 17 at 6:13

answered Mar 16 at 8:04

Paras KhoslaParas Khosla

2,643323

answered Mar 16 at 8:04

Paras KhoslaParas Khosla

2,643323

answered Mar 16 at 8:04

Paras KhoslaParas Khosla

2,643323

2

$begingroup$

I assume:
$$lim_x to 1^+ fracsqrtx^2-1+xcolorred-1 sqrt4x-4+x^2-1 = lim_x to 1^+ fracsqrtx-1cdot sqrtx+1+(sqrtx-1)^2 sqrt4x-4+(sqrtx-1)^2(x+1) =\
lim_x to 1^+ fracsqrtx-1cdot (sqrtx+1+sqrtx-1)sqrtx-1cdot (2+(sqrtx-1)(x+1)) =fracsqrt22.$$

share|cite|improve this answer

answered Mar 16 at 8:08

farruhotafarruhota

21.5k2842

$endgroup$

add a comment | 

2

$begingroup$

I assume:
$$lim_x to 1^+ fracsqrtx^2-1+xcolorred-1 sqrt4x-4+x^2-1 = lim_x to 1^+ fracsqrtx-1cdot sqrtx+1+(sqrtx-1)^2 sqrt4x-4+(sqrtx-1)^2(x+1) =\
lim_x to 1^+ fracsqrtx-1cdot (sqrtx+1+sqrtx-1)sqrtx-1cdot (2+(sqrtx-1)(x+1)) =fracsqrt22.$$

share|cite|improve this answer

answered Mar 16 at 8:08

farruhotafarruhota

21.5k2842

$endgroup$

add a comment | 

$begingroup$

I assume:
$$lim_x to 1^+ fracsqrtx^2-1+xcolorred-1 sqrt4x-4+x^2-1 = lim_x to 1^+ fracsqrtx-1cdot sqrtx+1+(sqrtx-1)^2 sqrt4x-4+(sqrtx-1)^2(x+1) =\
lim_x to 1^+ fracsqrtx-1cdot (sqrtx+1+sqrtx-1)sqrtx-1cdot (2+(sqrtx-1)(x+1)) =fracsqrt22.$$

share|cite|improve this answer

answered Mar 16 at 8:08

farruhotafarruhota

21.5k2842

$endgroup$

share|cite|improve this answer

answered Mar 16 at 8:08

farruhotafarruhota

21.5k2842

answered Mar 16 at 8:08

farruhotafarruhota

21.5k2842

answered Mar 16 at 8:08

farruhotafarruhota

21.5k2842

0

$begingroup$

Set $x-1=t^2$, which you can because the limit is for $xto1^+$. Then the limit becomes
$$
lim_tto0^+fractsqrtt^2+2+t^22t+t^2(t^2+2)=
lim_tto0^+fracsqrtt^2+2+t2+t(t^2+2)
$$

share|cite|improve this answer

answered Mar 16 at 11:39

egregegreg

185k1486206

$endgroup$

add a comment | 

0

$begingroup$

Set $x-1=t^2$, which you can because the limit is for $xto1^+$. Then the limit becomes
$$
lim_tto0^+fractsqrtt^2+2+t^22t+t^2(t^2+2)=
lim_tto0^+fracsqrtt^2+2+t2+t(t^2+2)
$$

share|cite|improve this answer

answered Mar 16 at 11:39

egregegreg

185k1486206

$endgroup$

add a comment | 

$begingroup$

Set $x-1=t^2$, which you can because the limit is for $xto1^+$. Then the limit becomes
$$
lim_tto0^+fractsqrtt^2+2+t^22t+t^2(t^2+2)=
lim_tto0^+fracsqrtt^2+2+t2+t(t^2+2)
$$

share|cite|improve this answer

answered Mar 16 at 11:39

egregegreg

185k1486206

$endgroup$

share|cite|improve this answer

answered Mar 16 at 11:39

egregegreg

185k1486206

answered Mar 16 at 11:39

egregegreg

185k1486206

answered Mar 16 at 11:39

egregegreg

185k1486206