calculate limit without L'Hospital's Rules
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I'm struggling with this limit. We did not learn L'Hospital's Rule, Taylor expansions, etc. yet. Should be solvable with plain old limit arithmetic:
$lim_limits{x to 1} (frac{3}{1-sqrt{x}}-frac{2}{1-sqrt[3]{x}})$
I tried combining the fractions with the common denominator, I tried multiplying each one by the conjugates, substituting different values as t.
Still haven't had my breakthrough yet.
Please help, thanks
limits limits-without-lhopital
edited Jan 24 at 10:58
Bernard
122k740116
asked Jan 24 at 10:38
ikoikoiaikoikoia
1399
$endgroup$
add a comment |
$begingroup$
I'm struggling with this limit. We did not learn L'Hospital's Rule, Taylor expansions, etc. yet. Should be solvable with plain old limit arithmetic:
$lim_limits{x to 1} (frac{3}{1-sqrt{x}}-frac{2}{1-sqrt[3]{x}})$
I tried combining the fractions with the common denominator, I tried multiplying each one by the conjugates, substituting different values as t.
Still haven't had my breakthrough yet.
Please help, thanks
limits limits-without-lhopital
edited Jan 24 at 10:58
Bernard
122k740116
asked Jan 24 at 10:38
ikoikoiaikoikoia
1399
$endgroup$
$begingroup$
See math.stackexchange.com/questions/529961/…
$endgroup$
– lab bhattacharjee
Jan 24 at 11:56
add a comment |
$begingroup$
I'm struggling with this limit. We did not learn L'Hospital's Rule, Taylor expansions, etc. yet. Should be solvable with plain old limit arithmetic:
$lim_limits{x to 1} (frac{3}{1-sqrt{x}}-frac{2}{1-sqrt[3]{x}})$
I tried combining the fractions with the common denominator, I tried multiplying each one by the conjugates, substituting different values as t.
Still haven't had my breakthrough yet.
Please help, thanks
limits limits-without-lhopital
edited Jan 24 at 10:58
Bernard
122k740116
asked Jan 24 at 10:38
ikoikoiaikoikoia
1399
$endgroup$
I'm struggling with this limit. We did not learn L'Hospital's Rule, Taylor expansions, etc. yet. Should be solvable with plain old limit arithmetic:
$lim_limits{x to 1} (frac{3}{1-sqrt{x}}-frac{2}{1-sqrt[3]{x}})$
I tried combining the fractions with the common denominator, I tried multiplying each one by the conjugates, substituting different values as t.
Still haven't had my breakthrough yet.
Please help, thanks
limits limits-without-lhopital
limits limits-without-lhopital
edited Jan 24 at 10:58
Bernard
122k740116
asked Jan 24 at 10:38
ikoikoiaikoikoia
1399
edited Jan 24 at 10:58
Bernard
122k740116
asked Jan 24 at 10:38
ikoikoiaikoikoia
1399
edited Jan 24 at 10:58
Bernard
122k740116
edited Jan 24 at 10:58
Bernard
122k740116
edited Jan 24 at 10:58
Bernard
122k740116
122k740116
asked Jan 24 at 10:38
ikoikoiaikoikoia
1399
asked Jan 24 at 10:38
ikoikoiaikoikoia
1399
asked Jan 24 at 10:38
ikoikoiaikoikoia
1399
1399
$begingroup$
See math.stackexchange.com/questions/529961/…
$endgroup$
– lab bhattacharjee
Jan 24 at 11:56
add a comment |
$begingroup$
See math.stackexchange.com/questions/529961/…
$endgroup$
– lab bhattacharjee
Jan 24 at 11:56
$begingroup$
See math.stackexchange.com/questions/529961/…
$endgroup$
– lab bhattacharjee
Jan 24 at 11:56
$begingroup$
See math.stackexchange.com/questions/529961/…
$endgroup$
– lab bhattacharjee
Jan 24 at 11:56
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
Hint: We can substitute $x=t^6$ to obtain
$$
lim_{tto 1}frac{3}{1-t^3}-frac{2}{1-t^2}=lim_{tto 1}frac{1}{1-t}left(frac{3}{t^2+t+1}-frac{2}{1+t}right).
$$
answered Jan 24 at 10:50
SongSong
16.6k11043
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add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint: We can substitute $x=t^6$ to obtain
$$
lim_{tto 1}frac{3}{1-t^3}-frac{2}{1-t^2}=lim_{tto 1}frac{1}{1-t}left(frac{3}{t^2+t+1}-frac{2}{1+t}right).
$$
answered Jan 24 at 10:50
SongSong
16.6k11043
$endgroup$
add a comment |
$begingroup$
Hint: We can substitute $x=t^6$ to obtain
$$
lim_{tto 1}frac{3}{1-t^3}-frac{2}{1-t^2}=lim_{tto 1}frac{1}{1-t}left(frac{3}{t^2+t+1}-frac{2}{1+t}right).
$$
answered Jan 24 at 10:50
SongSong
16.6k11043
$endgroup$
add a comment |
$begingroup$
Hint: We can substitute $x=t^6$ to obtain
$$
lim_{tto 1}frac{3}{1-t^3}-frac{2}{1-t^2}=lim_{tto 1}frac{1}{1-t}left(frac{3}{t^2+t+1}-frac{2}{1+t}right).
$$
answered Jan 24 at 10:50
SongSong
16.6k11043
$endgroup$
Hint: We can substitute $x=t^6$ to obtain
$$
lim_{tto 1}frac{3}{1-t^3}-frac{2}{1-t^2}=lim_{tto 1}frac{1}{1-t}left(frac{3}{t^2+t+1}-frac{2}{1+t}right).
$$
answered Jan 24 at 10:50
SongSong
16.6k11043
answered Jan 24 at 10:50
SongSong
16.6k11043
answered Jan 24 at 10:50
SongSong
16.6k11043
answered Jan 24 at 10:50
SongSong
16.6k11043
16.6k11043
add a comment |
add a comment |
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$begingroup$
See math.stackexchange.com/questions/529961/…
$endgroup$
– lab bhattacharjee
Jan 24 at 11:56