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Evaluate limit without using L'H rule

Is there any way to evaluate this limit without applying de l'Hôpital rule nor series expansion?Is it possible to evaluate this limit without series expansion and l'Hôpital's rule?Find this limit without L'hopital Rule : $lim_xrightarrow +inftyfracx(1+sin(x))x-sqrt(1+x^2)$Limits without L'Hopitals RuleEvaluating the given limit without using L'H rule.Evaluate the limit without using the L'Hôpital's ruleSolve this limit without using L'Hôpital's ruleLimit of a quotient without using L'Hopital ruleTrig limit without using l'hopital's ruleEvaluate limit without L'Hopital's rule

1

$begingroup$

I was revising teacher's notes about L'H rule and I came across this limit.

$$lim_x to fracpi2 fracx-fracpi2sqrt1-sin x$$

The teacher tried to evaluate without L'H to demonstrate L'H is necessary here but I can't explain the first passage he did:

$$lim_x to fracpi2 fracx-fracpi2sqrt1-sin x = lim_x to fracpi2 frac1frac-cos x2sqrt1-sin x$$

Can you please help me?

limits limits-without-lhopital

share|cite|improve this question

edited Mar 21 at 12:11

Maria Mazur

49.6k1361124

asked Mar 21 at 12:05

zcbzcb

222

$endgroup$

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  It seems like an application of L' Hospital Rule
  $endgroup$
  – Shubham Johri
  Mar 21 at 12:07
  

  
  
  
  

add a comment | 

1

$begingroup$

I was revising teacher's notes about L'H rule and I came across this limit.

$$lim_x to fracpi2 fracx-fracpi2sqrt1-sin x$$

The teacher tried to evaluate without L'H to demonstrate L'H is necessary here but I can't explain the first passage he did:

$$lim_x to fracpi2 fracx-fracpi2sqrt1-sin x = lim_x to fracpi2 frac1frac-cos x2sqrt1-sin x$$

Can you please help me?

limits limits-without-lhopital

share|cite|improve this question

edited Mar 21 at 12:11

Maria Mazur

49.6k1361124

asked Mar 21 at 12:05

zcbzcb

222

$endgroup$

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  It seems like an application of L' Hospital Rule
  $endgroup$
  – Shubham Johri
  Mar 21 at 12:07
  

  
  
  
  

add a comment | 

$begingroup$

I was revising teacher's notes about L'H rule and I came across this limit.

$$lim_x to fracpi2 fracx-fracpi2sqrt1-sin x$$

The teacher tried to evaluate without L'H to demonstrate L'H is necessary here but I can't explain the first passage he did:

$$lim_x to fracpi2 fracx-fracpi2sqrt1-sin x = lim_x to fracpi2 frac1frac-cos x2sqrt1-sin x$$

Can you please help me?

limits limits-without-lhopital

share|cite|improve this question

edited Mar 21 at 12:11

Maria Mazur

49.6k1361124

asked Mar 21 at 12:05

zcbzcb

222

$endgroup$

share|cite|improve this question

edited Mar 21 at 12:11

Maria Mazur

49.6k1361124

asked Mar 21 at 12:05

zcbzcb

222

share|cite|improve this question

edited Mar 21 at 12:11

Maria Mazur

49.6k1361124

asked Mar 21 at 12:05

zcbzcb

222

edited Mar 21 at 12:11

Maria Mazur

49.6k1361124

edited Mar 21 at 12:11

Maria Mazur

49.6k1361124

asked Mar 21 at 12:05

zcbzcb

222

asked Mar 21 at 12:05

zcbzcb

222

4 Answers
4

active

oldest

votes

1

$begingroup$

Hint

$dfracpi2-x=2y$

$$lim_yto0dfrac-2ysqrt1-cos2y=-sqrt2lim_...dfrac y$$

So, the limit doesn't exist

share|cite|improve this answer

answered Mar 21 at 12:37

lab bhattacharjeelab bhattacharjee

228k15158279

$endgroup$

add a comment | 

0

$begingroup$

Let $h=x-pi/2implies hto 0$ as $xto pi/2$ $$beginalignedlim_xto pi/2dfracx-pi/2sqrt1-sin x&=lim_hto 0dfrachsqrt1-cos hcdotdfracsqrt1+cos hsqrt1+cos h\&=lim_hto 0dfrachmidsin hmidcdotlim_hto 0sqrt1+cos hendaligned$$

Define $f(h)=dfrachsqrt1+cos hmid sin hmid$. Note that since $lim_hto 0^+f(h)ne lim_hto 0^-f(h)$, the limit does not exist.

share|cite|improve this answer

answered Mar 21 at 13:42

Paras KhoslaParas Khosla

2,792423

$endgroup$

add a comment | 

0

$begingroup$

Here is a way without L'Hospital using

• 
  $lim_yto 0fracsin yy = 1$ (which can be shown without L'Hospital) and


• $cos (y+ fracpi2) = -sin y$

$$fracx-fracpi2sqrt1-sin x= underbracefracx-fracpi2_stackrely = x - fracpi2= fracy
begincases
stackrely to 0^+longrightarrow & 1 \
stackrely to 0^-longrightarrow & -1 \
endcases
cdot sqrt1+sin x
begincases
stackrelx to (fracpi2)^+longrightarrow & sqrt2 \
stackrelx to (fracpi2)^-longrightarrow & -sqrt2 \
endcases$$

So, only the one-sided limits exist and are different.

share|cite|improve this answer

edited Mar 21 at 13:52

answered Mar 21 at 12:34

trancelocationtrancelocation

13.5k1828

$endgroup$

add a comment | 

0

$begingroup$

Hint: Put $t=x-pi over 2$, then you got

$$lim_t to 0 fractsqrt1-cos t$$

Now write $s=t/2$, so $t=2s$...

share|cite|improve this answer

edited Mar 21 at 18:51

answered Mar 21 at 12:09

Maria MazurMaria Mazur

49.6k1361124

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  You have a small typo... it is $1- cos t$.
  $endgroup$
  – PierreCarre
  Mar 21 at 12:17
  

  
  
  
  

add a comment | 

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1

$begingroup$

Hint

$dfracpi2-x=2y$

$$lim_yto0dfrac-2ysqrt1-cos2y=-sqrt2lim_...dfrac y$$

So, the limit doesn't exist

share|cite|improve this answer

answered Mar 21 at 12:37

lab bhattacharjeelab bhattacharjee

228k15158279

$endgroup$

add a comment | 

1

$begingroup$

Hint

$dfracpi2-x=2y$

$$lim_yto0dfrac-2ysqrt1-cos2y=-sqrt2lim_...dfrac y$$

So, the limit doesn't exist

share|cite|improve this answer

answered Mar 21 at 12:37

lab bhattacharjeelab bhattacharjee

228k15158279

$endgroup$

add a comment | 

$begingroup$

Hint

$dfracpi2-x=2y$

$$lim_yto0dfrac-2ysqrt1-cos2y=-sqrt2lim_...dfrac y$$

So, the limit doesn't exist

share|cite|improve this answer

answered Mar 21 at 12:37

lab bhattacharjeelab bhattacharjee

228k15158279

$endgroup$

share|cite|improve this answer

answered Mar 21 at 12:37

lab bhattacharjeelab bhattacharjee

228k15158279

answered Mar 21 at 12:37

lab bhattacharjeelab bhattacharjee

228k15158279

answered Mar 21 at 12:37

lab bhattacharjeelab bhattacharjee

228k15158279

0

$begingroup$

Let $h=x-pi/2implies hto 0$ as $xto pi/2$ $$beginalignedlim_xto pi/2dfracx-pi/2sqrt1-sin x&=lim_hto 0dfrachsqrt1-cos hcdotdfracsqrt1+cos hsqrt1+cos h\&=lim_hto 0dfrachmidsin hmidcdotlim_hto 0sqrt1+cos hendaligned$$

Define $f(h)=dfrachsqrt1+cos hmid sin hmid$. Note that since $lim_hto 0^+f(h)ne lim_hto 0^-f(h)$, the limit does not exist.

share|cite|improve this answer

answered Mar 21 at 13:42

Paras KhoslaParas Khosla

2,792423

$endgroup$

add a comment | 

0

$begingroup$

Let $h=x-pi/2implies hto 0$ as $xto pi/2$ $$beginalignedlim_xto pi/2dfracx-pi/2sqrt1-sin x&=lim_hto 0dfrachsqrt1-cos hcdotdfracsqrt1+cos hsqrt1+cos h\&=lim_hto 0dfrachmidsin hmidcdotlim_hto 0sqrt1+cos hendaligned$$

Define $f(h)=dfrachsqrt1+cos hmid sin hmid$. Note that since $lim_hto 0^+f(h)ne lim_hto 0^-f(h)$, the limit does not exist.

share|cite|improve this answer

answered Mar 21 at 13:42

Paras KhoslaParas Khosla

2,792423

$endgroup$

add a comment | 

$begingroup$

Let $h=x-pi/2implies hto 0$ as $xto pi/2$ $$beginalignedlim_xto pi/2dfracx-pi/2sqrt1-sin x&=lim_hto 0dfrachsqrt1-cos hcdotdfracsqrt1+cos hsqrt1+cos h\&=lim_hto 0dfrachmidsin hmidcdotlim_hto 0sqrt1+cos hendaligned$$

Define $f(h)=dfrachsqrt1+cos hmid sin hmid$. Note that since $lim_hto 0^+f(h)ne lim_hto 0^-f(h)$, the limit does not exist.

share|cite|improve this answer

answered Mar 21 at 13:42

Paras KhoslaParas Khosla

2,792423

$endgroup$

share|cite|improve this answer

answered Mar 21 at 13:42

Paras KhoslaParas Khosla

2,792423

answered Mar 21 at 13:42

Paras KhoslaParas Khosla

2,792423

answered Mar 21 at 13:42

Paras KhoslaParas Khosla

2,792423

0

$begingroup$

Here is a way without L'Hospital using

• 
  $lim_yto 0fracsin yy = 1$ (which can be shown without L'Hospital) and


• $cos (y+ fracpi2) = -sin y$

$$fracx-fracpi2sqrt1-sin x= underbracefracx-fracpi2_stackrely = x - fracpi2= fracy
begincases
stackrely to 0^+longrightarrow & 1 \
stackrely to 0^-longrightarrow & -1 \
endcases
cdot sqrt1+sin x
begincases
stackrelx to (fracpi2)^+longrightarrow & sqrt2 \
stackrelx to (fracpi2)^-longrightarrow & -sqrt2 \
endcases$$

So, only the one-sided limits exist and are different.

share|cite|improve this answer

edited Mar 21 at 13:52

answered Mar 21 at 12:34

trancelocationtrancelocation

13.5k1828

$endgroup$

add a comment | 

0

$begingroup$

Here is a way without L'Hospital using

• 
  $lim_yto 0fracsin yy = 1$ (which can be shown without L'Hospital) and


• $cos (y+ fracpi2) = -sin y$

$$fracx-fracpi2sqrt1-sin x= underbracefracx-fracpi2_stackrely = x - fracpi2= fracy
begincases
stackrely to 0^+longrightarrow & 1 \
stackrely to 0^-longrightarrow & -1 \
endcases
cdot sqrt1+sin x
begincases
stackrelx to (fracpi2)^+longrightarrow & sqrt2 \
stackrelx to (fracpi2)^-longrightarrow & -sqrt2 \
endcases$$

So, only the one-sided limits exist and are different.

share|cite|improve this answer

edited Mar 21 at 13:52

answered Mar 21 at 12:34

trancelocationtrancelocation

13.5k1828

$endgroup$

add a comment | 

$begingroup$

Here is a way without L'Hospital using

• 
  $lim_yto 0fracsin yy = 1$ (which can be shown without L'Hospital) and


• $cos (y+ fracpi2) = -sin y$

$$fracx-fracpi2sqrt1-sin x= underbracefracx-fracpi2_stackrely = x - fracpi2= fracy
begincases
stackrely to 0^+longrightarrow & 1 \
stackrely to 0^-longrightarrow & -1 \
endcases
cdot sqrt1+sin x
begincases
stackrelx to (fracpi2)^+longrightarrow & sqrt2 \
stackrelx to (fracpi2)^-longrightarrow & -sqrt2 \
endcases$$

So, only the one-sided limits exist and are different.

share|cite|improve this answer

edited Mar 21 at 13:52

answered Mar 21 at 12:34

trancelocationtrancelocation

13.5k1828

$endgroup$

share|cite|improve this answer

edited Mar 21 at 13:52

answered Mar 21 at 12:34

trancelocationtrancelocation

13.5k1828

answered Mar 21 at 12:34

trancelocationtrancelocation

13.5k1828

answered Mar 21 at 12:34

trancelocationtrancelocation

13.5k1828

0

$begingroup$

Hint: Put $t=x-pi over 2$, then you got

$$lim_t to 0 fractsqrt1-cos t$$

Now write $s=t/2$, so $t=2s$...

share|cite|improve this answer

edited Mar 21 at 18:51

answered Mar 21 at 12:09

Maria MazurMaria Mazur

49.6k1361124

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  You have a small typo... it is $1- cos t$.
  $endgroup$
  – PierreCarre
  Mar 21 at 12:17
  

  
  
  
  

add a comment | 

0

$begingroup$

Hint: Put $t=x-pi over 2$, then you got

$$lim_t to 0 fractsqrt1-cos t$$

Now write $s=t/2$, so $t=2s$...

share|cite|improve this answer

edited Mar 21 at 18:51

answered Mar 21 at 12:09

Maria MazurMaria Mazur

49.6k1361124

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  You have a small typo... it is $1- cos t$.
  $endgroup$
  – PierreCarre
  Mar 21 at 12:17
  

  
  
  
  

add a comment | 

$begingroup$

Hint: Put $t=x-pi over 2$, then you got

$$lim_t to 0 fractsqrt1-cos t$$

Now write $s=t/2$, so $t=2s$...

share|cite|improve this answer

edited Mar 21 at 18:51

answered Mar 21 at 12:09

Maria MazurMaria Mazur

49.6k1361124

$endgroup$

share|cite|improve this answer

edited Mar 21 at 18:51

answered Mar 21 at 12:09

Maria MazurMaria Mazur

49.6k1361124

answered Mar 21 at 12:09

Maria MazurMaria Mazur

49.6k1361124

answered Mar 21 at 12:09

Maria MazurMaria Mazur

49.6k1361124