Trigonometric Limit (No L'Hôpital) [duplicate]

This question already has an answer here:

$limlimits_{xto 0} frac{tan x - sin x}{x^3}$?

5 answers

I came across with this limit:

$lim_{x to 0}frac{tan x - sin x}{x^3}$

I started working it out this way:

$lim_{x to 0}(frac{tan x}{x}times frac{1}{x^2}) - (frac{sin x }{x}times frac{1}{x^2})$ = $lim_{x to 0}(1times frac{1}{x^2}) - (1times frac{1}{x^2})$ = $lim_{x to 0}frac{1}{x^2} - frac{1}{x^2}$ = $0$

This is wrong since the solution is $frac{1}{2}$. So my question is why can't I do this?

asked Nov 24 '18 at 16:22

Jakcjones

588

marked as duplicate by Community♦ Nov 24 '18 at 16:27

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

$begingroup$
Are you plugging 0 in for $x$ in the trigonometric functions in the second equality?
$endgroup$
– gd1035
Nov 24 '18 at 16:25

$begingroup$
This question has already been asked.
$endgroup$
– KM101
Nov 24 '18 at 16:25

$begingroup$
How did you conclude $$lim_{x to 0}frac{1}{x^2} - frac{1}{x^2} = 0 $$? . It is in $$infty - infty$$ form.
$endgroup$
– Chinmaya mishra
Nov 24 '18 at 16:26

$begingroup$
You are right, I'll delete it
$endgroup$
– Jakcjones
Nov 24 '18 at 16:27

add a comment |

This question already has an answer here:

$limlimits_{xto 0} frac{tan x - sin x}{x^3}$?

5 answers

I came across with this limit:

$lim_{x to 0}frac{tan x - sin x}{x^3}$

I started working it out this way:

$lim_{x to 0}(frac{tan x}{x}times frac{1}{x^2}) - (frac{sin x }{x}times frac{1}{x^2})$ = $lim_{x to 0}(1times frac{1}{x^2}) - (1times frac{1}{x^2})$ = $lim_{x to 0}frac{1}{x^2} - frac{1}{x^2}$ = $0$

This is wrong since the solution is $frac{1}{2}$. So my question is why can't I do this?

asked Nov 24 '18 at 16:22

Jakcjones

588

marked as duplicate by Community♦ Nov 24 '18 at 16:27

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

$begingroup$
Are you plugging 0 in for $x$ in the trigonometric functions in the second equality?
$endgroup$
– gd1035
Nov 24 '18 at 16:25

$begingroup$
This question has already been asked.
$endgroup$
– KM101
Nov 24 '18 at 16:25

$begingroup$
How did you conclude $$lim_{x to 0}frac{1}{x^2} - frac{1}{x^2} = 0 $$? . It is in $$infty - infty$$ form.
$endgroup$
– Chinmaya mishra
Nov 24 '18 at 16:26

$begingroup$
You are right, I'll delete it
$endgroup$
– Jakcjones
Nov 24 '18 at 16:27

add a comment |

This question already has an answer here:

$limlimits_{xto 0} frac{tan x - sin x}{x^3}$?

5 answers

I came across with this limit:

$lim_{x to 0}frac{tan x - sin x}{x^3}$

I started working it out this way:

$lim_{x to 0}(frac{tan x}{x}times frac{1}{x^2}) - (frac{sin x }{x}times frac{1}{x^2})$ = $lim_{x to 0}(1times frac{1}{x^2}) - (1times frac{1}{x^2})$ = $lim_{x to 0}frac{1}{x^2} - frac{1}{x^2}$ = $0$

This is wrong since the solution is $frac{1}{2}$. So my question is why can't I do this?

asked Nov 24 '18 at 16:22

Jakcjones

588

This question already has an answer here:

$limlimits_{xto 0} frac{tan x - sin x}{x^3}$?

5 answers

I came across with this limit:

$lim_{x to 0}frac{tan x - sin x}{x^3}$

I started working it out this way:

$lim_{x to 0}(frac{tan x}{x}times frac{1}{x^2}) - (frac{sin x }{x}times frac{1}{x^2})$ = $lim_{x to 0}(1times frac{1}{x^2}) - (1times frac{1}{x^2})$ = $lim_{x to 0}frac{1}{x^2} - frac{1}{x^2}$ = $0$

This is wrong since the solution is $frac{1}{2}$. So my question is why can't I do this?

This question already has an answer here:

$limlimits_{xto 0} frac{tan x - sin x}{x^3}$?

5 answers

calculus limits trigonometry limits-without-lhopital

asked Nov 24 '18 at 16:22

Jakcjones

588

asked Nov 24 '18 at 16:22

Jakcjones

588

asked Nov 24 '18 at 16:22

Jakcjones

588

asked Nov 24 '18 at 16:22

Jakcjones

588

asked Nov 24 '18 at 16:22

Jakcjones

588

marked as duplicate by Community♦ Nov 24 '18 at 16:27

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

marked as duplicate by Community♦ Nov 24 '18 at 16:27

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

$begingroup$
Are you plugging 0 in for $x$ in the trigonometric functions in the second equality?
$endgroup$
– gd1035
Nov 24 '18 at 16:25

$begingroup$
This question has already been asked.
$endgroup$
– KM101
Nov 24 '18 at 16:25

$begingroup$
How did you conclude $$lim_{x to 0}frac{1}{x^2} - frac{1}{x^2} = 0 $$? . It is in $$infty - infty$$ form.
$endgroup$
– Chinmaya mishra
Nov 24 '18 at 16:26

$begingroup$
You are right, I'll delete it
$endgroup$
– Jakcjones
Nov 24 '18 at 16:27

add a comment |

$begingroup$
Are you plugging 0 in for $x$ in the trigonometric functions in the second equality?
$endgroup$
– gd1035
Nov 24 '18 at 16:25

$begingroup$
This question has already been asked.
$endgroup$
– KM101
Nov 24 '18 at 16:25

$begingroup$
How did you conclude $$lim_{x to 0}frac{1}{x^2} - frac{1}{x^2} = 0 $$? . It is in $$infty - infty$$ form.
$endgroup$
– Chinmaya mishra
Nov 24 '18 at 16:26

$begingroup$
You are right, I'll delete it
$endgroup$
– Jakcjones
Nov 24 '18 at 16:27

Are you plugging 0 in for $x$ in the trigonometric functions in the second equality?

– gd1035
Nov 24 '18 at 16:25

This question has already been asked.

– KM101
Nov 24 '18 at 16:25

How did you conclude $$lim_{x to 0}frac{1}{x^2} - frac{1}{x^2} = 0 $$? . It is in $$infty - infty$$ form.

– Chinmaya mishra
Nov 24 '18 at 16:26

You are right, I'll delete it

– Jakcjones
Nov 24 '18 at 16:27

add a comment |

1 Answer
1

active

oldest

votes

$$frac{tan x-sin x}{x^3}=frac{sin x}xcdotfrac{frac1{cos x}-1}{x^2}=frac{sin x}xcdotfrac1{cos x}cdotfrac{1-cos x}{x^2}=$$

$$=frac{sin x}xcdotfrac1{cos x}cdotfrac{sin^2x}{x^2}cdotfrac1{1+cos x}xrightarrow[xto0]{}1cdot1cdot1^2cdotfrac1{1+1}=frac12$$

answered Nov 24 '18 at 16:26

DonAntonio

177k1492225

add a comment |

1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

$$frac{tan x-sin x}{x^3}=frac{sin x}xcdotfrac{frac1{cos x}-1}{x^2}=frac{sin x}xcdotfrac1{cos x}cdotfrac{1-cos x}{x^2}=$$

$$=frac{sin x}xcdotfrac1{cos x}cdotfrac{sin^2x}{x^2}cdotfrac1{1+cos x}xrightarrow[xto0]{}1cdot1cdot1^2cdotfrac1{1+1}=frac12$$

answered Nov 24 '18 at 16:26

DonAntonio

177k1492225

add a comment |

$$frac{tan x-sin x}{x^3}=frac{sin x}xcdotfrac{frac1{cos x}-1}{x^2}=frac{sin x}xcdotfrac1{cos x}cdotfrac{1-cos x}{x^2}=$$

$$=frac{sin x}xcdotfrac1{cos x}cdotfrac{sin^2x}{x^2}cdotfrac1{1+cos x}xrightarrow[xto0]{}1cdot1cdot1^2cdotfrac1{1+1}=frac12$$

answered Nov 24 '18 at 16:26

DonAntonio

177k1492225

add a comment |

$$frac{tan x-sin x}{x^3}=frac{sin x}xcdotfrac{frac1{cos x}-1}{x^2}=frac{sin x}xcdotfrac1{cos x}cdotfrac{1-cos x}{x^2}=$$

$$=frac{sin x}xcdotfrac1{cos x}cdotfrac{sin^2x}{x^2}cdotfrac1{1+cos x}xrightarrow[xto0]{}1cdot1cdot1^2cdotfrac1{1+1}=frac12$$

answered Nov 24 '18 at 16:26

DonAntonio

177k1492225

$$frac{tan x-sin x}{x^3}=frac{sin x}xcdotfrac{frac1{cos x}-1}{x^2}=frac{sin x}xcdotfrac1{cos x}cdotfrac{1-cos x}{x^2}=$$

$$=frac{sin x}xcdotfrac1{cos x}cdotfrac{sin^2x}{x^2}cdotfrac1{1+cos x}xrightarrow[xto0]{}1cdot1cdot1^2cdotfrac1{1+1}=frac12$$

answered Nov 24 '18 at 16:26

DonAntonio

177k1492225

answered Nov 24 '18 at 16:26

DonAntonio

177k1492225

answered Nov 24 '18 at 16:26

DonAntonio

177k1492225

answered Nov 24 '18 at 16:26

DonAntonio

177k1492225

add a comment |

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