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Advance Calculus Limit question

2

1

$begingroup$

I'm trying to compute this limit without the use of L'Hopital's rule:

$$lim_x to 0^+ frac4^-1/x+4^1/x4^-1/x-4^1/x$$

I've been trying to multiply by the lcd and doing other creative stuff... anyone have any suggestions on theorems or techniques?

calculus

share|cite|improve this question

edited Apr 2 at 23:56

Foobaz John

23k41552

asked Apr 1 at 23:49

Kevin CalderonKevin Calderon

513

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Why have you changed the title of the OP?
  $endgroup$
  – Paras Khosla
  Apr 2 at 7:24
  

  
  
  
  

add a comment | 

2

1

$begingroup$

I'm trying to compute this limit without the use of L'Hopital's rule:

$$lim_x to 0^+ frac4^-1/x+4^1/x4^-1/x-4^1/x$$

I've been trying to multiply by the lcd and doing other creative stuff... anyone have any suggestions on theorems or techniques?

calculus

share|cite|improve this question

edited Apr 2 at 23:56

Foobaz John

23k41552

asked Apr 1 at 23:49

Kevin CalderonKevin Calderon

513

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Why have you changed the title of the OP?
  $endgroup$
  – Paras Khosla
  Apr 2 at 7:24
  

  
  
  
  

add a comment | 

$begingroup$

I'm trying to compute this limit without the use of L'Hopital's rule:

$$lim_x to 0^+ frac4^-1/x+4^1/x4^-1/x-4^1/x$$

I've been trying to multiply by the lcd and doing other creative stuff... anyone have any suggestions on theorems or techniques?

calculus

share|cite|improve this question

edited Apr 2 at 23:56

Foobaz John

23k41552

asked Apr 1 at 23:49

Kevin CalderonKevin Calderon

513

$endgroup$

share|cite|improve this question

edited Apr 2 at 23:56

Foobaz John

23k41552

asked Apr 1 at 23:49

Kevin CalderonKevin Calderon

513

share|cite|improve this question

edited Apr 2 at 23:56

Foobaz John

23k41552

asked Apr 1 at 23:49

Kevin CalderonKevin Calderon

513

edited Apr 2 at 23:56

Foobaz John

23k41552

edited Apr 2 at 23:56

Foobaz John

23k41552

asked Apr 1 at 23:49

Kevin CalderonKevin Calderon

513

asked Apr 1 at 23:49

Kevin CalderonKevin Calderon

513

3 Answers
3

active

oldest

votes

6

$begingroup$

Write the limit as
$$
lim_xto 0+frac1+4^-2/x-1+4^-2/x
$$
and use the fact that
$$
lim_xto 0+frac-2x=-infty.
$$
to find that the limit equals $-1$.

share|cite|improve this answer

answered Apr 1 at 23:57

Foobaz JohnFoobaz John

23k41552

$endgroup$

add a comment | 

3

$begingroup$

A substitution can be helpful, as it transforms the expression into a rational function:

• Set $y=4^frac1x$ and consider $y to +infty$
  

begineqnarray* frac4^-1/x+4^1/x4^-1/x-4^1/x
& stackrely=4^frac1x= & fracfrac1y+yfrac1y-y \
& = & fracfrac1y^2+1frac1y^2-1 \
& stackrely to +inftylongrightarrow & frac0+10-1 = -1
endeqnarray*

share|cite|improve this answer

answered Apr 2 at 4:11

trancelocationtrancelocation

14.5k1929

$endgroup$

add a comment | 

0

$begingroup$

$$lim_xto 0^+dfrac4^-1/x+4^1/x4^-1/x-4^1/x=lim_xto 0^+dfrac4^-2/x+14^-2/x-1$$

Clearly as $xto 0^+$, $2/xto infty$. Since the power of $4$ is $-2/x$, it must go to $0$. Effectively we have $frac0+10-1=-1$. Hence the required limit is $-1$.

share|cite|improve this answer

answered Apr 2 at 5:25

Paras KhoslaParas Khosla

3,2851627

$endgroup$

add a comment | 

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6

$begingroup$

Write the limit as
$$
lim_xto 0+frac1+4^-2/x-1+4^-2/x
$$
and use the fact that
$$
lim_xto 0+frac-2x=-infty.
$$
to find that the limit equals $-1$.

share|cite|improve this answer

answered Apr 1 at 23:57

Foobaz JohnFoobaz John

23k41552

$endgroup$

add a comment | 

6

$begingroup$

Write the limit as
$$
lim_xto 0+frac1+4^-2/x-1+4^-2/x
$$
and use the fact that
$$
lim_xto 0+frac-2x=-infty.
$$
to find that the limit equals $-1$.

share|cite|improve this answer

answered Apr 1 at 23:57

Foobaz JohnFoobaz John

23k41552

$endgroup$

add a comment | 

$begingroup$

Write the limit as
$$
lim_xto 0+frac1+4^-2/x-1+4^-2/x
$$
and use the fact that
$$
lim_xto 0+frac-2x=-infty.
$$
to find that the limit equals $-1$.

share|cite|improve this answer

answered Apr 1 at 23:57

Foobaz JohnFoobaz John

23k41552

$endgroup$

share|cite|improve this answer

answered Apr 1 at 23:57

Foobaz JohnFoobaz John

23k41552

answered Apr 1 at 23:57

Foobaz JohnFoobaz John

23k41552

answered Apr 1 at 23:57

Foobaz JohnFoobaz John

23k41552

3

$begingroup$

A substitution can be helpful, as it transforms the expression into a rational function:

• Set $y=4^frac1x$ and consider $y to +infty$
  

begineqnarray* frac4^-1/x+4^1/x4^-1/x-4^1/x
& stackrely=4^frac1x= & fracfrac1y+yfrac1y-y \
& = & fracfrac1y^2+1frac1y^2-1 \
& stackrely to +inftylongrightarrow & frac0+10-1 = -1
endeqnarray*

share|cite|improve this answer

answered Apr 2 at 4:11

trancelocationtrancelocation

14.5k1929

$endgroup$

add a comment | 

3

$begingroup$

A substitution can be helpful, as it transforms the expression into a rational function:

• Set $y=4^frac1x$ and consider $y to +infty$
  

begineqnarray* frac4^-1/x+4^1/x4^-1/x-4^1/x
& stackrely=4^frac1x= & fracfrac1y+yfrac1y-y \
& = & fracfrac1y^2+1frac1y^2-1 \
& stackrely to +inftylongrightarrow & frac0+10-1 = -1
endeqnarray*

share|cite|improve this answer

answered Apr 2 at 4:11

trancelocationtrancelocation

14.5k1929

$endgroup$

add a comment | 

$begingroup$

A substitution can be helpful, as it transforms the expression into a rational function:

• Set $y=4^frac1x$ and consider $y to +infty$
  

begineqnarray* frac4^-1/x+4^1/x4^-1/x-4^1/x
& stackrely=4^frac1x= & fracfrac1y+yfrac1y-y \
& = & fracfrac1y^2+1frac1y^2-1 \
& stackrely to +inftylongrightarrow & frac0+10-1 = -1
endeqnarray*

share|cite|improve this answer

answered Apr 2 at 4:11

trancelocationtrancelocation

14.5k1929

$endgroup$

share|cite|improve this answer

answered Apr 2 at 4:11

trancelocationtrancelocation

14.5k1929

answered Apr 2 at 4:11

trancelocationtrancelocation

14.5k1929

answered Apr 2 at 4:11

trancelocationtrancelocation

14.5k1929

0

$begingroup$

$$lim_xto 0^+dfrac4^-1/x+4^1/x4^-1/x-4^1/x=lim_xto 0^+dfrac4^-2/x+14^-2/x-1$$

Clearly as $xto 0^+$, $2/xto infty$. Since the power of $4$ is $-2/x$, it must go to $0$. Effectively we have $frac0+10-1=-1$. Hence the required limit is $-1$.

share|cite|improve this answer

answered Apr 2 at 5:25

Paras KhoslaParas Khosla

3,2851627

$endgroup$

add a comment | 

0

$begingroup$

$$lim_xto 0^+dfrac4^-1/x+4^1/x4^-1/x-4^1/x=lim_xto 0^+dfrac4^-2/x+14^-2/x-1$$

Clearly as $xto 0^+$, $2/xto infty$. Since the power of $4$ is $-2/x$, it must go to $0$. Effectively we have $frac0+10-1=-1$. Hence the required limit is $-1$.

share|cite|improve this answer

answered Apr 2 at 5:25

Paras KhoslaParas Khosla

3,2851627

$endgroup$

add a comment | 

$begingroup$

$$lim_xto 0^+dfrac4^-1/x+4^1/x4^-1/x-4^1/x=lim_xto 0^+dfrac4^-2/x+14^-2/x-1$$

Clearly as $xto 0^+$, $2/xto infty$. Since the power of $4$ is $-2/x$, it must go to $0$. Effectively we have $frac0+10-1=-1$. Hence the required limit is $-1$.

share|cite|improve this answer

answered Apr 2 at 5:25

Paras KhoslaParas Khosla

3,2851627

$endgroup$

share|cite|improve this answer

answered Apr 2 at 5:25

Paras KhoslaParas Khosla

3,2851627

answered Apr 2 at 5:25

Paras KhoslaParas Khosla

3,2851627

answered Apr 2 at 5:25

Paras KhoslaParas Khosla

3,2851627