Calculate the limit without l'Hopital ruleHow to calculate limit without L'HopitalSolving limit without...
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Calculate the limit without l'Hopital rule
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Calculate the limit without l'Hopital rule
How to calculate limit without L'HopitalSolving limit without L'Hopitalwithout using l'hopital ruleSolving limit of radicals without L'Hopital $lim_{xto 64} frac{sqrt x - 8}{sqrt[3] x - 4} $Calculate limit without L'Hopital's ruleCalculate the limit without using L'Hopital's RuleSolving limit without L'Hopital ruleHow to calculate this limit without L'Hopital rule?Calculate the following limit without L'Hopital
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$begingroup$
I have the following limit:
$$
lim_{xto 7}dfrac{x^2-4x-21}{x-4-sqrt{x+2}}
$$
I could easily calculate the limit = 12 using the l'Hopital rule.
Could you please suggest any other ways to solve this limit without using the l'Hopital rule?
Thank you
limits limits-without-lhopital
asked 8 hours ago
Freshman42Freshman42
3401 silver badge14 bronze badges
$endgroup$
add a comment
|
$begingroup$
I have the following limit:
$$
lim_{xto 7}dfrac{x^2-4x-21}{x-4-sqrt{x+2}}
$$
I could easily calculate the limit = 12 using the l'Hopital rule.
Could you please suggest any other ways to solve this limit without using the l'Hopital rule?
Thank you
limits limits-without-lhopital
asked 8 hours ago
Freshman42Freshman42
3401 silver badge14 bronze badges
$endgroup$
1
$begingroup$
@Freshman42 My guess is that you multiplied out the numerator. There's no reason to do that. Instead, you should note $$frac{(x^2 - 4x - 21)(x-4+sqrt{x+2})}{(x-4-sqrt{x+2})(x-4+sqrt{x+2})} = frac{(x^2 - 4x - 21)(x-4+sqrt{x+2})}{x^2-9x+14}$$ Now factor the two quadratics.
$endgroup$
– Brian Moehring
8 hours ago
$begingroup$
@BrianMoehring: Thank you it works!
$endgroup$
– Freshman42
8 hours ago
add a comment
|
$begingroup$
I have the following limit:
$$
lim_{xto 7}dfrac{x^2-4x-21}{x-4-sqrt{x+2}}
$$
I could easily calculate the limit = 12 using the l'Hopital rule.
Could you please suggest any other ways to solve this limit without using the l'Hopital rule?
Thank you
limits limits-without-lhopital
asked 8 hours ago
Freshman42Freshman42
3401 silver badge14 bronze badges
$endgroup$
I have the following limit:
$$
lim_{xto 7}dfrac{x^2-4x-21}{x-4-sqrt{x+2}}
$$
I could easily calculate the limit = 12 using the l'Hopital rule.
Could you please suggest any other ways to solve this limit without using the l'Hopital rule?
Thank you
limits limits-without-lhopital
limits limits-without-lhopital
asked 8 hours ago
Freshman42Freshman42
3401 silver badge14 bronze badges
asked 8 hours ago
Freshman42Freshman42
3401 silver badge14 bronze badges
asked 8 hours ago
Freshman42Freshman42
3401 silver badge14 bronze badges
asked 8 hours ago
Freshman42Freshman42
3401 silver badge14 bronze badges
asked 8 hours ago
Freshman42Freshman42
3401 silver badge14 bronze badges
3401 silver badge14 bronze badges
1
$begingroup$
@Freshman42 My guess is that you multiplied out the numerator. There's no reason to do that. Instead, you should note $$frac{(x^2 - 4x - 21)(x-4+sqrt{x+2})}{(x-4-sqrt{x+2})(x-4+sqrt{x+2})} = frac{(x^2 - 4x - 21)(x-4+sqrt{x+2})}{x^2-9x+14}$$ Now factor the two quadratics.
$endgroup$
– Brian Moehring
8 hours ago
$begingroup$
@BrianMoehring: Thank you it works!
$endgroup$
– Freshman42
8 hours ago
add a comment
|
1
$begingroup$
@Freshman42 My guess is that you multiplied out the numerator. There's no reason to do that. Instead, you should note $$frac{(x^2 - 4x - 21)(x-4+sqrt{x+2})}{(x-4-sqrt{x+2})(x-4+sqrt{x+2})} = frac{(x^2 - 4x - 21)(x-4+sqrt{x+2})}{x^2-9x+14}$$ Now factor the two quadratics.
$endgroup$
– Brian Moehring
8 hours ago
$begingroup$
@BrianMoehring: Thank you it works!
$endgroup$
– Freshman42
8 hours ago
1
1
$begingroup$
@Freshman42 My guess is that you multiplied out the numerator. There's no reason to do that. Instead, you should note $$frac{(x^2 - 4x - 21)(x-4+sqrt{x+2})}{(x-4-sqrt{x+2})(x-4+sqrt{x+2})} = frac{(x^2 - 4x - 21)(x-4+sqrt{x+2})}{x^2-9x+14}$$ Now factor the two quadratics.
$endgroup$
– Brian Moehring
8 hours ago
$begingroup$
@Freshman42 My guess is that you multiplied out the numerator. There's no reason to do that. Instead, you should note $$frac{(x^2 - 4x - 21)(x-4+sqrt{x+2})}{(x-4-sqrt{x+2})(x-4+sqrt{x+2})} = frac{(x^2 - 4x - 21)(x-4+sqrt{x+2})}{x^2-9x+14}$$ Now factor the two quadratics.
$endgroup$
– Brian Moehring
8 hours ago
$begingroup$
@BrianMoehring: Thank you it works!
$endgroup$
– Freshman42
8 hours ago
$begingroup$
@BrianMoehring: Thank you it works!
$endgroup$
– Freshman42
8 hours ago
add a comment
|
3 Answers
3
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oldest
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$begingroup$
An alternative to @MatthewDaly's comment: write $y:=sqrt{x+2}$ so you want$$lim_{yto3}frac{y^4-8y^2-9}{y^2-y-6}=lim_{yto3}frac{y^3+3y^2+y+3}{y+2}=frac{3^3+3times 3^2+3+3}{5}=12.$$
answered 8 hours ago
J.G.J.G.
47k2 gold badges42 silver badges62 bronze badges
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$$lim_{x→7}frac{x^2−4x−21}{x−4−sqrt{x+2}}cdotfrac{x−4+sqrt{x+2}}{x−4+sqrt{x+2}}\
=lim_{x→7}frac{(x^2−4x−21)cdot(x−4+sqrt{x+2})}{(x−4)^2−(x+2)}\
=lim_{x→7}frac{(x^2−4x−21)cdot(x−4+sqrt{x+2})}{x^2-9x+14}\
=lim_{x→7}frac{(x+3)cdot(x−4+sqrt{x+2})}{x-2}\
=frac{10cdot6}{5}=12$$
since $(x-7)$ can be factored out of both of those quadratics.
answered 8 hours ago
Matthew DalyMatthew Daly
4,9721 gold badge7 silver badges27 bronze badges
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The numerator can be factored as $(x-7)(x+3)$. Now consider
$$
lim_{xto7}frac{x-4-sqrt{x+2}}{x-7}=lim_{xto7}frac{x-7-(sqrt{x+2}-3)}{x-7}=
1-lim_{xto7}frac{x+2-9}{(x-7)(sqrt{x+2}+3)}=1-frac{1}{6}=frac{5}{6}
$$
So your limit is
$$
lim_{xto7}frac{x-7}{x-4-sqrt{x+2}}(x+3)=frac{6}{5}cdot10=12
$$
answered 8 hours ago
egregegreg
192k14 gold badges91 silver badges218 bronze badges
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
An alternative to @MatthewDaly's comment: write $y:=sqrt{x+2}$ so you want$$lim_{yto3}frac{y^4-8y^2-9}{y^2-y-6}=lim_{yto3}frac{y^3+3y^2+y+3}{y+2}=frac{3^3+3times 3^2+3+3}{5}=12.$$
answered 8 hours ago
J.G.J.G.
47k2 gold badges42 silver badges62 bronze badges
$endgroup$
add a comment
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$begingroup$
An alternative to @MatthewDaly's comment: write $y:=sqrt{x+2}$ so you want$$lim_{yto3}frac{y^4-8y^2-9}{y^2-y-6}=lim_{yto3}frac{y^3+3y^2+y+3}{y+2}=frac{3^3+3times 3^2+3+3}{5}=12.$$
answered 8 hours ago
J.G.J.G.
47k2 gold badges42 silver badges62 bronze badges
$endgroup$
add a comment
|
$begingroup$
An alternative to @MatthewDaly's comment: write $y:=sqrt{x+2}$ so you want$$lim_{yto3}frac{y^4-8y^2-9}{y^2-y-6}=lim_{yto3}frac{y^3+3y^2+y+3}{y+2}=frac{3^3+3times 3^2+3+3}{5}=12.$$
answered 8 hours ago
J.G.J.G.
47k2 gold badges42 silver badges62 bronze badges
$endgroup$
An alternative to @MatthewDaly's comment: write $y:=sqrt{x+2}$ so you want$$lim_{yto3}frac{y^4-8y^2-9}{y^2-y-6}=lim_{yto3}frac{y^3+3y^2+y+3}{y+2}=frac{3^3+3times 3^2+3+3}{5}=12.$$
answered 8 hours ago
J.G.J.G.
47k2 gold badges42 silver badges62 bronze badges
answered 8 hours ago
J.G.J.G.
47k2 gold badges42 silver badges62 bronze badges
answered 8 hours ago
J.G.J.G.
47k2 gold badges42 silver badges62 bronze badges
answered 8 hours ago
J.G.J.G.
47k2 gold badges42 silver badges62 bronze badges
47k2 gold badges42 silver badges62 bronze badges
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$begingroup$
$$lim_{x→7}frac{x^2−4x−21}{x−4−sqrt{x+2}}cdotfrac{x−4+sqrt{x+2}}{x−4+sqrt{x+2}}\
=lim_{x→7}frac{(x^2−4x−21)cdot(x−4+sqrt{x+2})}{(x−4)^2−(x+2)}\
=lim_{x→7}frac{(x^2−4x−21)cdot(x−4+sqrt{x+2})}{x^2-9x+14}\
=lim_{x→7}frac{(x+3)cdot(x−4+sqrt{x+2})}{x-2}\
=frac{10cdot6}{5}=12$$
since $(x-7)$ can be factored out of both of those quadratics.
answered 8 hours ago
Matthew DalyMatthew Daly
4,9721 gold badge7 silver badges27 bronze badges
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$begingroup$
$$lim_{x→7}frac{x^2−4x−21}{x−4−sqrt{x+2}}cdotfrac{x−4+sqrt{x+2}}{x−4+sqrt{x+2}}\
=lim_{x→7}frac{(x^2−4x−21)cdot(x−4+sqrt{x+2})}{(x−4)^2−(x+2)}\
=lim_{x→7}frac{(x^2−4x−21)cdot(x−4+sqrt{x+2})}{x^2-9x+14}\
=lim_{x→7}frac{(x+3)cdot(x−4+sqrt{x+2})}{x-2}\
=frac{10cdot6}{5}=12$$
since $(x-7)$ can be factored out of both of those quadratics.
answered 8 hours ago
Matthew DalyMatthew Daly
4,9721 gold badge7 silver badges27 bronze badges
$endgroup$
add a comment
|
$begingroup$
$$lim_{x→7}frac{x^2−4x−21}{x−4−sqrt{x+2}}cdotfrac{x−4+sqrt{x+2}}{x−4+sqrt{x+2}}\
=lim_{x→7}frac{(x^2−4x−21)cdot(x−4+sqrt{x+2})}{(x−4)^2−(x+2)}\
=lim_{x→7}frac{(x^2−4x−21)cdot(x−4+sqrt{x+2})}{x^2-9x+14}\
=lim_{x→7}frac{(x+3)cdot(x−4+sqrt{x+2})}{x-2}\
=frac{10cdot6}{5}=12$$
since $(x-7)$ can be factored out of both of those quadratics.
answered 8 hours ago
Matthew DalyMatthew Daly
4,9721 gold badge7 silver badges27 bronze badges
$endgroup$
$$lim_{x→7}frac{x^2−4x−21}{x−4−sqrt{x+2}}cdotfrac{x−4+sqrt{x+2}}{x−4+sqrt{x+2}}\
=lim_{x→7}frac{(x^2−4x−21)cdot(x−4+sqrt{x+2})}{(x−4)^2−(x+2)}\
=lim_{x→7}frac{(x^2−4x−21)cdot(x−4+sqrt{x+2})}{x^2-9x+14}\
=lim_{x→7}frac{(x+3)cdot(x−4+sqrt{x+2})}{x-2}\
=frac{10cdot6}{5}=12$$
since $(x-7)$ can be factored out of both of those quadratics.
answered 8 hours ago
Matthew DalyMatthew Daly
4,9721 gold badge7 silver badges27 bronze badges
answered 8 hours ago
Matthew DalyMatthew Daly
4,9721 gold badge7 silver badges27 bronze badges
answered 8 hours ago
Matthew DalyMatthew Daly
4,9721 gold badge7 silver badges27 bronze badges
answered 8 hours ago
Matthew DalyMatthew Daly
4,9721 gold badge7 silver badges27 bronze badges
4,9721 gold badge7 silver badges27 bronze badges
add a comment
|
add a comment
|
$begingroup$
The numerator can be factored as $(x-7)(x+3)$. Now consider
$$
lim_{xto7}frac{x-4-sqrt{x+2}}{x-7}=lim_{xto7}frac{x-7-(sqrt{x+2}-3)}{x-7}=
1-lim_{xto7}frac{x+2-9}{(x-7)(sqrt{x+2}+3)}=1-frac{1}{6}=frac{5}{6}
$$
So your limit is
$$
lim_{xto7}frac{x-7}{x-4-sqrt{x+2}}(x+3)=frac{6}{5}cdot10=12
$$
answered 8 hours ago
egregegreg
192k14 gold badges91 silver badges218 bronze badges
$endgroup$
add a comment
|
$begingroup$
The numerator can be factored as $(x-7)(x+3)$. Now consider
$$
lim_{xto7}frac{x-4-sqrt{x+2}}{x-7}=lim_{xto7}frac{x-7-(sqrt{x+2}-3)}{x-7}=
1-lim_{xto7}frac{x+2-9}{(x-7)(sqrt{x+2}+3)}=1-frac{1}{6}=frac{5}{6}
$$
So your limit is
$$
lim_{xto7}frac{x-7}{x-4-sqrt{x+2}}(x+3)=frac{6}{5}cdot10=12
$$
answered 8 hours ago
egregegreg
192k14 gold badges91 silver badges218 bronze badges
$endgroup$
add a comment
|
$begingroup$
The numerator can be factored as $(x-7)(x+3)$. Now consider
$$
lim_{xto7}frac{x-4-sqrt{x+2}}{x-7}=lim_{xto7}frac{x-7-(sqrt{x+2}-3)}{x-7}=
1-lim_{xto7}frac{x+2-9}{(x-7)(sqrt{x+2}+3)}=1-frac{1}{6}=frac{5}{6}
$$
So your limit is
$$
lim_{xto7}frac{x-7}{x-4-sqrt{x+2}}(x+3)=frac{6}{5}cdot10=12
$$
answered 8 hours ago
egregegreg
192k14 gold badges91 silver badges218 bronze badges
$endgroup$
The numerator can be factored as $(x-7)(x+3)$. Now consider
$$
lim_{xto7}frac{x-4-sqrt{x+2}}{x-7}=lim_{xto7}frac{x-7-(sqrt{x+2}-3)}{x-7}=
1-lim_{xto7}frac{x+2-9}{(x-7)(sqrt{x+2}+3)}=1-frac{1}{6}=frac{5}{6}
$$
So your limit is
$$
lim_{xto7}frac{x-7}{x-4-sqrt{x+2}}(x+3)=frac{6}{5}cdot10=12
$$
answered 8 hours ago
egregegreg
192k14 gold badges91 silver badges218 bronze badges
answered 8 hours ago
egregegreg
192k14 gold badges91 silver badges218 bronze badges
answered 8 hours ago
egregegreg
192k14 gold badges91 silver badges218 bronze badges
answered 8 hours ago
egregegreg
192k14 gold badges91 silver badges218 bronze badges
192k14 gold badges91 silver badges218 bronze badges
add a comment
|
add a comment
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$begingroup$
@Freshman42 My guess is that you multiplied out the numerator. There's no reason to do that. Instead, you should note $$frac{(x^2 - 4x - 21)(x-4+sqrt{x+2})}{(x-4-sqrt{x+2})(x-4+sqrt{x+2})} = frac{(x^2 - 4x - 21)(x-4+sqrt{x+2})}{x^2-9x+14}$$ Now factor the two quadratics.
$endgroup$
– Brian Moehring
8 hours ago
$begingroup$
@BrianMoehring: Thank you it works!
$endgroup$
– Freshman42
8 hours ago