Trigonometric Limit (No L'Hôpital) [duplicate]
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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 '18 at 16:22
JakcjonesJakcjones
588
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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.
add a comment |
$begingroup$
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 '18 at 16:22
JakcjonesJakcjones
588
$endgroup$
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.
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Are you plugging 0 in for $x$ in the trigonometric functions in the second equality?
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– gd1035
Nov 24 '18 at 16:25
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This question has already been asked.
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– KM101
Nov 24 '18 at 16:25
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How did you conclude $$lim_{x to 0}frac{1}{x^2} - frac{1}{x^2} = 0 $$? . It is in $$infty - infty$$ form.
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– Chinmaya mishra
Nov 24 '18 at 16:26
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You are right, I'll delete it
$endgroup$
– Jakcjones
Nov 24 '18 at 16:27
add a comment |
$begingroup$
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 '18 at 16:22
JakcjonesJakcjones
588
$endgroup$
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 '18 at 16:22
JakcjonesJakcjones
588
asked Nov 24 '18 at 16:22
JakcjonesJakcjones
588
asked Nov 24 '18 at 16:22
JakcjonesJakcjones
588
asked Nov 24 '18 at 16:22
JakcjonesJakcjones
588
asked Nov 24 '18 at 16:22
JakcjonesJakcjones
588
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
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This question has already been asked.
$endgroup$
– KM101
Nov 24 '18 at 16:25
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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
$begingroup$
Are you plugging 0 in for $x$ in the trigonometric functions in the second equality?
$endgroup$
– gd1035
Nov 24 '18 at 16:25
$begingroup$
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
$begingroup$
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
$begingroup$
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
$begingroup$
You are right, I'll delete it
$endgroup$
– Jakcjones
Nov 24 '18 at 16:27
add a comment |
1 Answer
1
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$$frac{tan x-sin x}{x^3}=frac{sin x}xcdotfrac{frac1{cos x}-1}{x^2}=frac{sin x}xcdotfrac1{cos x}cdotfrac{1-cos x}{x^2}=$$
$$=frac{sin x}xcdotfrac1{cos x}cdotfrac{sin^2x}{x^2}cdotfrac1{1+cos x}xrightarrow[xto0]{}1cdot1cdot1^2cdotfrac1{1+1}=frac12$$
answered Nov 24 '18 at 16:26
DonAntonioDonAntonio
177k1492225
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add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
$$frac{tan x-sin x}{x^3}=frac{sin x}xcdotfrac{frac1{cos x}-1}{x^2}=frac{sin x}xcdotfrac1{cos x}cdotfrac{1-cos x}{x^2}=$$
$$=frac{sin x}xcdotfrac1{cos x}cdotfrac{sin^2x}{x^2}cdotfrac1{1+cos x}xrightarrow[xto0]{}1cdot1cdot1^2cdotfrac1{1+1}=frac12$$
answered Nov 24 '18 at 16:26
DonAntonioDonAntonio
177k1492225
$endgroup$
add a comment |
$begingroup$
$$frac{tan x-sin x}{x^3}=frac{sin x}xcdotfrac{frac1{cos x}-1}{x^2}=frac{sin x}xcdotfrac1{cos x}cdotfrac{1-cos x}{x^2}=$$
$$=frac{sin x}xcdotfrac1{cos x}cdotfrac{sin^2x}{x^2}cdotfrac1{1+cos x}xrightarrow[xto0]{}1cdot1cdot1^2cdotfrac1{1+1}=frac12$$
answered Nov 24 '18 at 16:26
DonAntonioDonAntonio
177k1492225
$endgroup$
add a comment |
$begingroup$
$$frac{tan x-sin x}{x^3}=frac{sin x}xcdotfrac{frac1{cos x}-1}{x^2}=frac{sin x}xcdotfrac1{cos x}cdotfrac{1-cos x}{x^2}=$$
$$=frac{sin x}xcdotfrac1{cos x}cdotfrac{sin^2x}{x^2}cdotfrac1{1+cos x}xrightarrow[xto0]{}1cdot1cdot1^2cdotfrac1{1+1}=frac12$$
answered Nov 24 '18 at 16:26
DonAntonioDonAntonio
177k1492225
$endgroup$
$$frac{tan x-sin x}{x^3}=frac{sin x}xcdotfrac{frac1{cos x}-1}{x^2}=frac{sin x}xcdotfrac1{cos x}cdotfrac{1-cos x}{x^2}=$$
$$=frac{sin x}xcdotfrac1{cos x}cdotfrac{sin^2x}{x^2}cdotfrac1{1+cos x}xrightarrow[xto0]{}1cdot1cdot1^2cdotfrac1{1+1}=frac12$$
answered Nov 24 '18 at 16:26
DonAntonioDonAntonio
177k1492225
answered Nov 24 '18 at 16:26
DonAntonioDonAntonio
177k1492225
answered Nov 24 '18 at 16:26
DonAntonioDonAntonio
177k1492225
answered Nov 24 '18 at 16:26
DonAntonioDonAntonio
177k1492225
177k1492225
add a comment |
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