Trigonometric Limit (No L'Hôpital) [duplicate]
up vote
0
down vote
favorite
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?
calculus limits trigonometry limits-without-lhopital
asked Nov 24 at 16:22
Jakcjones
318
marked as duplicate by Community♦ Nov 24 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.
add a comment |
up vote
0
down vote
favorite
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?
calculus limits trigonometry limits-without-lhopital
asked Nov 24 at 16:22
Jakcjones
318
marked as duplicate by Community♦ Nov 24 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.
Are you plugging 0 in for $x$ in the trigonometric functions in the second equality?
– gd1035
Nov 24 at 16:25
This question has already been asked.
– KM101
Nov 24 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 at 16:26
You are right, I'll delete it
– Jakcjones
Nov 24 at 16:27
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
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?
calculus limits trigonometry limits-without-lhopital
asked Nov 24 at 16:22
Jakcjones
318
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
calculus limits trigonometry limits-without-lhopital
asked Nov 24 at 16:22
Jakcjones
318
asked Nov 24 at 16:22
Jakcjones
318
asked Nov 24 at 16:22
Jakcjones
318
asked Nov 24 at 16:22
Jakcjones
318
asked Nov 24 at 16:22
Jakcjones
318
318
marked as duplicate by Community♦ Nov 24 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 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.
Are you plugging 0 in for $x$ in the trigonometric functions in the second equality?
– gd1035
Nov 24 at 16:25
This question has already been asked.
– KM101
Nov 24 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 at 16:26
You are right, I'll delete it
– Jakcjones
Nov 24 at 16:27
add a comment |
Are you plugging 0 in for $x$ in the trigonometric functions in the second equality?
– gd1035
Nov 24 at 16:25
This question has already been asked.
– KM101
Nov 24 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 at 16:26
You are right, I'll delete it
– Jakcjones
Nov 24 at 16:27
Are you plugging 0 in for $x$ in the trigonometric functions in the second equality?
– gd1035
Nov 24 at 16:25
Are you plugging 0 in for $x$ in the trigonometric functions in the second equality?
– gd1035
Nov 24 at 16:25
This question has already been asked.
– KM101
Nov 24 at 16:25
This question has already been asked.
– KM101
Nov 24 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 at 16:26
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 at 16:26
You are right, I'll delete it
– Jakcjones
Nov 24 at 16:27
You are right, I'll delete it
– Jakcjones
Nov 24 at 16:27
add a comment |
1 Answer
1
active
oldest
votes
up vote
1
down vote
$$frac{tan x-sin x}{x^3}=frac{sin x}xcdotfrac{frac1{cos x}-1}{x^2}=frac{sin x}xcdotfrac1{cos x}cdotfrac{1-cos x}{x^2}=$$
$$=frac{sin x}xcdotfrac1{cos x}cdotfrac{sin^2x}{x^2}cdotfrac1{1+cos x}xrightarrow[xto0]{}1cdot1cdot1^2cdotfrac1{1+1}=frac12$$
answered Nov 24 at 16:26
DonAntonio
176k1491225
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
$$frac{tan x-sin x}{x^3}=frac{sin x}xcdotfrac{frac1{cos x}-1}{x^2}=frac{sin x}xcdotfrac1{cos x}cdotfrac{1-cos x}{x^2}=$$
$$=frac{sin x}xcdotfrac1{cos x}cdotfrac{sin^2x}{x^2}cdotfrac1{1+cos x}xrightarrow[xto0]{}1cdot1cdot1^2cdotfrac1{1+1}=frac12$$
answered Nov 24 at 16:26
DonAntonio
176k1491225
add a comment |
up vote
1
down vote
$$frac{tan x-sin x}{x^3}=frac{sin x}xcdotfrac{frac1{cos x}-1}{x^2}=frac{sin x}xcdotfrac1{cos x}cdotfrac{1-cos x}{x^2}=$$
$$=frac{sin x}xcdotfrac1{cos x}cdotfrac{sin^2x}{x^2}cdotfrac1{1+cos x}xrightarrow[xto0]{}1cdot1cdot1^2cdotfrac1{1+1}=frac12$$
answered Nov 24 at 16:26
DonAntonio
176k1491225
add a comment |
up vote
1
down vote
up vote
1
down vote
$$frac{tan x-sin x}{x^3}=frac{sin x}xcdotfrac{frac1{cos x}-1}{x^2}=frac{sin x}xcdotfrac1{cos x}cdotfrac{1-cos x}{x^2}=$$
$$=frac{sin x}xcdotfrac1{cos x}cdotfrac{sin^2x}{x^2}cdotfrac1{1+cos x}xrightarrow[xto0]{}1cdot1cdot1^2cdotfrac1{1+1}=frac12$$
answered Nov 24 at 16:26
DonAntonio
176k1491225
$$frac{tan x-sin x}{x^3}=frac{sin x}xcdotfrac{frac1{cos x}-1}{x^2}=frac{sin x}xcdotfrac1{cos x}cdotfrac{1-cos x}{x^2}=$$
$$=frac{sin x}xcdotfrac1{cos x}cdotfrac{sin^2x}{x^2}cdotfrac1{1+cos x}xrightarrow[xto0]{}1cdot1cdot1^2cdotfrac1{1+1}=frac12$$
answered Nov 24 at 16:26
DonAntonio
176k1491225
answered Nov 24 at 16:26
DonAntonio
176k1491225
answered Nov 24 at 16:26
DonAntonio
176k1491225
answered Nov 24 at 16:26
DonAntonio
176k1491225
176k1491225
add a comment |
add a comment |
Are you plugging 0 in for $x$ in the trigonometric functions in the second equality?
– gd1035
Nov 24 at 16:25
This question has already been asked.
– KM101
Nov 24 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 at 16:26
You are right, I'll delete it
– Jakcjones
Nov 24 at 16:27