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Computing the limit with L'Hôpital's Rule

0

$begingroup$

Problem:

$lim_{xto infty} frac{3n^2 - n^2 + n - 1}{1 - n + n^3 - 2n^5}$

What I tried

1. $lim_{xto infty} frac{4n + 1}{- 1 + 3n^2 - 10n^4}$


2. $lim_{xto infty} frac{4}{6n - 40n^3}$

I'm not sure what I'm supposed to do after this, any help would be much appreciated.

calculus limits

share|cite|improve this question

edited Mar 26 at 6:20

J. M. is a poor mathematician

61.3k5152291

asked Mar 26 at 0:10

user430574

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Based on your last four questions, I'd bet $100 that you're attending the University of Auckland.
  $endgroup$
  – Theo Bendit
  Mar 26 at 4:04
  

  
  
  
  

add a comment | 

0

$begingroup$

Problem:

$lim_{xto infty} frac{3n^2 - n^2 + n - 1}{1 - n + n^3 - 2n^5}$

What I tried

1. $lim_{xto infty} frac{4n + 1}{- 1 + 3n^2 - 10n^4}$


2. $lim_{xto infty} frac{4}{6n - 40n^3}$

I'm not sure what I'm supposed to do after this, any help would be much appreciated.

calculus limits

share|cite|improve this question

edited Mar 26 at 6:20

J. M. is a poor mathematician

61.3k5152291

asked Mar 26 at 0:10

user430574

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Based on your last four questions, I'd bet $100 that you're attending the University of Auckland.
  $endgroup$
  – Theo Bendit
  Mar 26 at 4:04
  

  
  
  
  

add a comment | 

$begingroup$

Problem:

$lim_{xto infty} frac{3n^2 - n^2 + n - 1}{1 - n + n^3 - 2n^5}$

What I tried

1. $lim_{xto infty} frac{4n + 1}{- 1 + 3n^2 - 10n^4}$


2. $lim_{xto infty} frac{4}{6n - 40n^3}$

I'm not sure what I'm supposed to do after this, any help would be much appreciated.

calculus limits

share|cite|improve this question

edited Mar 26 at 6:20

J. M. is a poor mathematician

61.3k5152291

asked Mar 26 at 0:10

user430574

$endgroup$

share|cite|improve this question

edited Mar 26 at 6:20

J. M. is a poor mathematician

61.3k5152291

asked Mar 26 at 0:10

user430574

share|cite|improve this question

edited Mar 26 at 6:20

J. M. is a poor mathematician

61.3k5152291

asked Mar 26 at 0:10

user430574

edited Mar 26 at 6:20

J. M. is a poor mathematician

61.3k5152291

edited Mar 26 at 6:20

J. M. is a poor mathematician

61.3k5152291

2 Answers
2

active

oldest

votes

3

$begingroup$

The limit is $0$ because $6n-40n^{3} to -infty$. [ $6n-40n^{3}= -n^{3}(40-frac 6 {n^{2}})$].

share|cite|improve this answer

answered Mar 26 at 0:16

Kavi Rama MurthyKavi Rama Murthy

76.4k53370

$endgroup$

add a comment | 

0

$begingroup$

Applying L'Hopital's rule twice gives us the expression in the form $-1/infty to 0$.

$$begin{aligned}lim_{color{red}{n}to infty}dfrac{3n^2-n^2+n-1}{1-n+n^3-2n^5}&=lim_{nto infty}dfrac{6n-2n+1}{-1+3n^2-10n^4}\&=lim_{nto infty}dfrac{4}{6n-40n^3}to 0end{aligned}$$

---

Aliter:

Dividing numerator and denominator by the highest power prevailing in the expression is usually another way to find limits of rational expressions especially when the variable tends to $infty$.

$$begin{aligned}lim_{color{red}{n}to infty}dfrac{3n^2-n^2+n-1}{1-n+n^3-2n^5}&=lim_{nto infty}dfrac{frac{3}{n^3}-frac{1}{n^3}+frac{1}{n^4}-frac{1}{n^5}}{frac{1}{n^5}-frac{1}{n^4}+frac{1}{n^2}-2}to 0end{aligned}$$

share|cite|improve this answer

answered Mar 26 at 5:49

Paras KhoslaParas Khosla

3,3121627

$endgroup$

add a comment | 

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3

$begingroup$

The limit is $0$ because $6n-40n^{3} to -infty$. [ $6n-40n^{3}= -n^{3}(40-frac 6 {n^{2}})$].

share|cite|improve this answer

answered Mar 26 at 0:16

Kavi Rama MurthyKavi Rama Murthy

76.4k53370

$endgroup$

add a comment | 

3

$begingroup$

The limit is $0$ because $6n-40n^{3} to -infty$. [ $6n-40n^{3}= -n^{3}(40-frac 6 {n^{2}})$].

share|cite|improve this answer

answered Mar 26 at 0:16

Kavi Rama MurthyKavi Rama Murthy

76.4k53370

$endgroup$

add a comment | 

$begingroup$

The limit is $0$ because $6n-40n^{3} to -infty$. [ $6n-40n^{3}= -n^{3}(40-frac 6 {n^{2}})$].

share|cite|improve this answer

answered Mar 26 at 0:16

Kavi Rama MurthyKavi Rama Murthy

76.4k53370

$endgroup$

share|cite|improve this answer

answered Mar 26 at 0:16

Kavi Rama MurthyKavi Rama Murthy

76.4k53370

answered Mar 26 at 0:16

Kavi Rama MurthyKavi Rama Murthy

76.4k53370

answered Mar 26 at 0:16

Kavi Rama MurthyKavi Rama Murthy

76.4k53370

0

$begingroup$

Applying L'Hopital's rule twice gives us the expression in the form $-1/infty to 0$.

$$begin{aligned}lim_{color{red}{n}to infty}dfrac{3n^2-n^2+n-1}{1-n+n^3-2n^5}&=lim_{nto infty}dfrac{6n-2n+1}{-1+3n^2-10n^4}\&=lim_{nto infty}dfrac{4}{6n-40n^3}to 0end{aligned}$$

---

Aliter:

Dividing numerator and denominator by the highest power prevailing in the expression is usually another way to find limits of rational expressions especially when the variable tends to $infty$.

$$begin{aligned}lim_{color{red}{n}to infty}dfrac{3n^2-n^2+n-1}{1-n+n^3-2n^5}&=lim_{nto infty}dfrac{frac{3}{n^3}-frac{1}{n^3}+frac{1}{n^4}-frac{1}{n^5}}{frac{1}{n^5}-frac{1}{n^4}+frac{1}{n^2}-2}to 0end{aligned}$$

share|cite|improve this answer

answered Mar 26 at 5:49

Paras KhoslaParas Khosla

3,3121627

$endgroup$

add a comment | 

0

$begingroup$

Applying L'Hopital's rule twice gives us the expression in the form $-1/infty to 0$.

$$begin{aligned}lim_{color{red}{n}to infty}dfrac{3n^2-n^2+n-1}{1-n+n^3-2n^5}&=lim_{nto infty}dfrac{6n-2n+1}{-1+3n^2-10n^4}\&=lim_{nto infty}dfrac{4}{6n-40n^3}to 0end{aligned}$$

---

Aliter:

Dividing numerator and denominator by the highest power prevailing in the expression is usually another way to find limits of rational expressions especially when the variable tends to $infty$.

$$begin{aligned}lim_{color{red}{n}to infty}dfrac{3n^2-n^2+n-1}{1-n+n^3-2n^5}&=lim_{nto infty}dfrac{frac{3}{n^3}-frac{1}{n^3}+frac{1}{n^4}-frac{1}{n^5}}{frac{1}{n^5}-frac{1}{n^4}+frac{1}{n^2}-2}to 0end{aligned}$$

share|cite|improve this answer

answered Mar 26 at 5:49

Paras KhoslaParas Khosla

3,3121627

$endgroup$

add a comment | 

$begingroup$

Applying L'Hopital's rule twice gives us the expression in the form $-1/infty to 0$.

$$begin{aligned}lim_{color{red}{n}to infty}dfrac{3n^2-n^2+n-1}{1-n+n^3-2n^5}&=lim_{nto infty}dfrac{6n-2n+1}{-1+3n^2-10n^4}\&=lim_{nto infty}dfrac{4}{6n-40n^3}to 0end{aligned}$$

---

Aliter:

Dividing numerator and denominator by the highest power prevailing in the expression is usually another way to find limits of rational expressions especially when the variable tends to $infty$.

$$begin{aligned}lim_{color{red}{n}to infty}dfrac{3n^2-n^2+n-1}{1-n+n^3-2n^5}&=lim_{nto infty}dfrac{frac{3}{n^3}-frac{1}{n^3}+frac{1}{n^4}-frac{1}{n^5}}{frac{1}{n^5}-frac{1}{n^4}+frac{1}{n^2}-2}to 0end{aligned}$$

share|cite|improve this answer

answered Mar 26 at 5:49

Paras KhoslaParas Khosla

3,3121627

$endgroup$

share|cite|improve this answer

answered Mar 26 at 5:49

Paras KhoslaParas Khosla

3,3121627

answered Mar 26 at 5:49

Paras KhoslaParas Khosla

3,3121627

answered Mar 26 at 5:49

Paras KhoslaParas Khosla

3,3121627