Mixing
Evaluating the limit $lim_{x to infty}left(sqrt{x^4 -x^3+1}-sqrt{x^4+15x^2-5},right)$
I wanna know how to do this limit
$lim_{x to infty}left(sqrt{x^4 -x^3+1}-sqrt{x^4+15x^2-5},right)$
calculus limits infinity
asked Nov 22 '18 at 4:32
Franco CabreraFranco Cabrera
64
add a comment |
I wanna know how to do this limit
$lim_{x to infty}left(sqrt{x^4 -x^3+1}-sqrt{x^4+15x^2-5},right)$
calculus limits infinity
asked Nov 22 '18 at 4:32
Franco CabreraFranco Cabrera
64
Try multiplying and dividing by the conjugate root to eliminate the roots.
– D.B.
Nov 22 '18 at 4:37
thats what i did, but i got stuck
– Franco Cabrera
Nov 22 '18 at 4:41
math.stackexchange.com/questions/2959619/…
– lab bhattacharjee
Nov 22 '18 at 4:41
Remember that intuitively, when you work with limits of polynomials, you can often just replace the polynomial with it's leading term and get the right answer. In this case you get $0$
– Ovi
Nov 22 '18 at 4:49
add a comment |
I wanna know how to do this limit
$lim_{x to infty}left(sqrt{x^4 -x^3+1}-sqrt{x^4+15x^2-5},right)$
calculus limits infinity
asked Nov 22 '18 at 4:32
Franco CabreraFranco Cabrera
64
I wanna know how to do this limit
$lim_{x to infty}left(sqrt{x^4 -x^3+1}-sqrt{x^4+15x^2-5},right)$
calculus limits infinity
calculus limits infinity
asked Nov 22 '18 at 4:32
Franco CabreraFranco Cabrera
64
asked Nov 22 '18 at 4:32
Franco CabreraFranco Cabrera
64
edited Nov 22 '18 at 4:37
Franco Cabrera
asked Nov 22 '18 at 4:32
Franco CabreraFranco Cabrera
64
asked Nov 22 '18 at 4:32
Franco CabreraFranco Cabrera
64
asked Nov 22 '18 at 4:32
Franco CabreraFranco Cabrera
64
64
Try multiplying and dividing by the conjugate root to eliminate the roots.
– D.B.
Nov 22 '18 at 4:37
thats what i did, but i got stuck
– Franco Cabrera
Nov 22 '18 at 4:41
math.stackexchange.com/questions/2959619/…
– lab bhattacharjee
Nov 22 '18 at 4:41
Remember that intuitively, when you work with limits of polynomials, you can often just replace the polynomial with it's leading term and get the right answer. In this case you get $0$
– Ovi
Nov 22 '18 at 4:49
add a comment |
Try multiplying and dividing by the conjugate root to eliminate the roots.
– D.B.
Nov 22 '18 at 4:37
thats what i did, but i got stuck
– Franco Cabrera
Nov 22 '18 at 4:41
math.stackexchange.com/questions/2959619/…
– lab bhattacharjee
Nov 22 '18 at 4:41
Remember that intuitively, when you work with limits of polynomials, you can often just replace the polynomial with it's leading term and get the right answer. In this case you get $0$
– Ovi
Nov 22 '18 at 4:49
Try multiplying and dividing by the conjugate root to eliminate the roots.
– D.B.
Nov 22 '18 at 4:37
Try multiplying and dividing by the conjugate root to eliminate the roots.
– D.B.
Nov 22 '18 at 4:37
thats what i did, but i got stuck
– Franco Cabrera
Nov 22 '18 at 4:41
thats what i did, but i got stuck
– Franco Cabrera
Nov 22 '18 at 4:41
math.stackexchange.com/questions/2959619/…
– lab bhattacharjee
Nov 22 '18 at 4:41
math.stackexchange.com/questions/2959619/…
– lab bhattacharjee
Nov 22 '18 at 4:41
Remember that intuitively, when you work with limits of polynomials, you can often just replace the polynomial with it's leading term and get the right answer. In this case you get $0$
– Ovi
Nov 22 '18 at 4:49
Remember that intuitively, when you work with limits of polynomials, you can often just replace the polynomial with it's leading term and get the right answer. In this case you get $0$
– Ovi
Nov 22 '18 at 4:49
add a comment |
3 Answers
3
active
oldest
votes
First of all, rationalise the expression by multiplying numerator and denominator by$displaystyle sqrt{x^{4} -x^{3} +1} + sqrt{x^{4} +15x^{2} -5}$
begin{equation*}
end{equation*}
We get
begin{gather*}
lim xrightarrow infty frac{-x^{3} -15x^{2} +6}{sqrt{x^{4} -x^{3} +1} + sqrt{x^{4} +15x^{2} -5}}\
end{gather*}
Divide numerator and denominator by $displaystyle x^{2}$
begin{equation*}
lim xrightarrow infty frac{-x-15+frac{6}{x^{2}}}{sqrt{1-frac{1}{x} +frac{1}{x^{4}}} +sqrt{1+frac{15}{x^{2}} -frac{5}{x^{4}}}}
end{equation*}
Clearly the given expression tends to$displaystyle -infty $ as $displaystyle xrightarrow infty $
begin{equation*}
end{equation*}
answered Nov 22 '18 at 5:49
Dikshit GautamDikshit Gautam
795
add a comment |
You should multiply by the conjugate of the expression, then proceed.
begin{align}
frac{left(sqrt{x^4 -x^3+1}-sqrt{x^4+15x^2-5},right)left(sqrt{x^4 -x^3+1}+sqrt{x^4+15x^2-5},right)}{left(sqrt{x^4 -x^3+1}+sqrt{x^4+15x^2-5},right)}\
end{align}
begin{align}
frac{(x^4 -x^3+1)-(x^4+15x^2-5)}{left(sqrt{x^4 -x^3+1}+sqrt{x^4+15x^2-5},right)}
end{align}
Operate and calculate the limit.
edited Nov 22 '18 at 6:03
b00n heT
10.2k12234
answered Nov 22 '18 at 4:48
Alberto Torrejon ValenzuelaAlberto Torrejon Valenzuela
276
negative! that's cool!
– Christopher Marley
Nov 22 '18 at 4:52
What does that mean? Sorry I`m not a fully native english speaker so those two words in the same sentence make me confuse.
– Alberto Torrejon Valenzuela
Nov 22 '18 at 4:54
and then, should i evaluate de sign of de function to see the lateral limits?
– Franco Cabrera
Nov 22 '18 at 4:54
No need if you just want to give a result of the limit. If you want to study the function then I suppose you should
– Alberto Torrejon Valenzuela
Nov 22 '18 at 4:56
Alberto, despues de eso tendria que evaluar el signo de f? ver si los limites laterales son iguales y obtener su valor?
– Franco Cabrera
Nov 22 '18 at 4:56
|
show 3 more comments
Another manipulation:
Let $a,b >0$, real.
$a-b=(√a-√b)(√a+√b)$, then
$√a-√b = dfrac{a-b}{√a+√b}.$
$a: = (x^4-x^3+1)^{1/2},$
$b=(x^4+15x^2-5)^{1/2}.$
$small{f(x):=}$
$small {=dfrac{-x^3-15x^2+6}{(x^4-x^3+1)^{1/2}+(x^4+15x^2-5)^{1/2}}}$
Note : For large x (positive) the numerator is negative, the denominator positive: Increasing the denominator makes the fraction bigger.
Hence
$f(x) lt $
$dfrac{-x^3+6}{(x^4+1)^{1/2}+(x^4+15x^2)^{1/2}} lt $
$dfrac{-x^3+6}{√2x^2+√2x^2}=$
$dfrac{-x^3+6}{(2√2)x^2}.$
Take the limit.
answered Nov 22 '18 at 9:34
Peter SzilasPeter Szilas
10.8k2720
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
First of all, rationalise the expression by multiplying numerator and denominator by$displaystyle sqrt{x^{4} -x^{3} +1} + sqrt{x^{4} +15x^{2} -5}$
begin{equation*}
end{equation*}
We get
begin{gather*}
lim xrightarrow infty frac{-x^{3} -15x^{2} +6}{sqrt{x^{4} -x^{3} +1} + sqrt{x^{4} +15x^{2} -5}}\
end{gather*}
Divide numerator and denominator by $displaystyle x^{2}$
begin{equation*}
lim xrightarrow infty frac{-x-15+frac{6}{x^{2}}}{sqrt{1-frac{1}{x} +frac{1}{x^{4}}} +sqrt{1+frac{15}{x^{2}} -frac{5}{x^{4}}}}
end{equation*}
Clearly the given expression tends to$displaystyle -infty $ as $displaystyle xrightarrow infty $
begin{equation*}
end{equation*}
answered Nov 22 '18 at 5:49
Dikshit GautamDikshit Gautam
795
add a comment |
First of all, rationalise the expression by multiplying numerator and denominator by$displaystyle sqrt{x^{4} -x^{3} +1} + sqrt{x^{4} +15x^{2} -5}$
begin{equation*}
end{equation*}
We get
begin{gather*}
lim xrightarrow infty frac{-x^{3} -15x^{2} +6}{sqrt{x^{4} -x^{3} +1} + sqrt{x^{4} +15x^{2} -5}}\
end{gather*}
Divide numerator and denominator by $displaystyle x^{2}$
begin{equation*}
lim xrightarrow infty frac{-x-15+frac{6}{x^{2}}}{sqrt{1-frac{1}{x} +frac{1}{x^{4}}} +sqrt{1+frac{15}{x^{2}} -frac{5}{x^{4}}}}
end{equation*}
Clearly the given expression tends to$displaystyle -infty $ as $displaystyle xrightarrow infty $
begin{equation*}
end{equation*}
answered Nov 22 '18 at 5:49
Dikshit GautamDikshit Gautam
795
add a comment |
First of all, rationalise the expression by multiplying numerator and denominator by$displaystyle sqrt{x^{4} -x^{3} +1} + sqrt{x^{4} +15x^{2} -5}$
begin{equation*}
end{equation*}
We get
begin{gather*}
lim xrightarrow infty frac{-x^{3} -15x^{2} +6}{sqrt{x^{4} -x^{3} +1} + sqrt{x^{4} +15x^{2} -5}}\
end{gather*}
Divide numerator and denominator by $displaystyle x^{2}$
begin{equation*}
lim xrightarrow infty frac{-x-15+frac{6}{x^{2}}}{sqrt{1-frac{1}{x} +frac{1}{x^{4}}} +sqrt{1+frac{15}{x^{2}} -frac{5}{x^{4}}}}
end{equation*}
Clearly the given expression tends to$displaystyle -infty $ as $displaystyle xrightarrow infty $
begin{equation*}
end{equation*}
answered Nov 22 '18 at 5:49
Dikshit GautamDikshit Gautam
795
First of all, rationalise the expression by multiplying numerator and denominator by$displaystyle sqrt{x^{4} -x^{3} +1} + sqrt{x^{4} +15x^{2} -5}$
begin{equation*}
end{equation*}
We get
begin{gather*}
lim xrightarrow infty frac{-x^{3} -15x^{2} +6}{sqrt{x^{4} -x^{3} +1} + sqrt{x^{4} +15x^{2} -5}}\
end{gather*}
Divide numerator and denominator by $displaystyle x^{2}$
begin{equation*}
lim xrightarrow infty frac{-x-15+frac{6}{x^{2}}}{sqrt{1-frac{1}{x} +frac{1}{x^{4}}} +sqrt{1+frac{15}{x^{2}} -frac{5}{x^{4}}}}
end{equation*}
Clearly the given expression tends to$displaystyle -infty $ as $displaystyle xrightarrow infty $
begin{equation*}
end{equation*}
answered Nov 22 '18 at 5:49
Dikshit GautamDikshit Gautam
795
answered Nov 22 '18 at 5:49
Dikshit GautamDikshit Gautam
795
answered Nov 22 '18 at 5:49
Dikshit GautamDikshit Gautam
795
answered Nov 22 '18 at 5:49
Dikshit GautamDikshit Gautam
795
795
add a comment |
add a comment |
You should multiply by the conjugate of the expression, then proceed.
begin{align}
frac{left(sqrt{x^4 -x^3+1}-sqrt{x^4+15x^2-5},right)left(sqrt{x^4 -x^3+1}+sqrt{x^4+15x^2-5},right)}{left(sqrt{x^4 -x^3+1}+sqrt{x^4+15x^2-5},right)}\
end{align}
begin{align}
frac{(x^4 -x^3+1)-(x^4+15x^2-5)}{left(sqrt{x^4 -x^3+1}+sqrt{x^4+15x^2-5},right)}
end{align}
Operate and calculate the limit.
edited Nov 22 '18 at 6:03
b00n heT
10.2k12234
answered Nov 22 '18 at 4:48
Alberto Torrejon ValenzuelaAlberto Torrejon Valenzuela
276
negative! that's cool!
– Christopher Marley
Nov 22 '18 at 4:52
What does that mean? Sorry I`m not a fully native english speaker so those two words in the same sentence make me confuse.
– Alberto Torrejon Valenzuela
Nov 22 '18 at 4:54
and then, should i evaluate de sign of de function to see the lateral limits?
– Franco Cabrera
Nov 22 '18 at 4:54
No need if you just want to give a result of the limit. If you want to study the function then I suppose you should
– Alberto Torrejon Valenzuela
Nov 22 '18 at 4:56
Alberto, despues de eso tendria que evaluar el signo de f? ver si los limites laterales son iguales y obtener su valor?
– Franco Cabrera
Nov 22 '18 at 4:56
|
show 3 more comments
You should multiply by the conjugate of the expression, then proceed.
begin{align}
frac{left(sqrt{x^4 -x^3+1}-sqrt{x^4+15x^2-5},right)left(sqrt{x^4 -x^3+1}+sqrt{x^4+15x^2-5},right)}{left(sqrt{x^4 -x^3+1}+sqrt{x^4+15x^2-5},right)}\
end{align}
begin{align}
frac{(x^4 -x^3+1)-(x^4+15x^2-5)}{left(sqrt{x^4 -x^3+1}+sqrt{x^4+15x^2-5},right)}
end{align}
Operate and calculate the limit.
edited Nov 22 '18 at 6:03
b00n heT
10.2k12234
answered Nov 22 '18 at 4:48
Alberto Torrejon ValenzuelaAlberto Torrejon Valenzuela
276
negative! that's cool!
– Christopher Marley
Nov 22 '18 at 4:52
What does that mean? Sorry I`m not a fully native english speaker so those two words in the same sentence make me confuse.
– Alberto Torrejon Valenzuela
Nov 22 '18 at 4:54
and then, should i evaluate de sign of de function to see the lateral limits?
– Franco Cabrera
Nov 22 '18 at 4:54
No need if you just want to give a result of the limit. If you want to study the function then I suppose you should
– Alberto Torrejon Valenzuela
Nov 22 '18 at 4:56
Alberto, despues de eso tendria que evaluar el signo de f? ver si los limites laterales son iguales y obtener su valor?
– Franco Cabrera
Nov 22 '18 at 4:56
|
show 3 more comments
You should multiply by the conjugate of the expression, then proceed.
begin{align}
frac{left(sqrt{x^4 -x^3+1}-sqrt{x^4+15x^2-5},right)left(sqrt{x^4 -x^3+1}+sqrt{x^4+15x^2-5},right)}{left(sqrt{x^4 -x^3+1}+sqrt{x^4+15x^2-5},right)}\
end{align}
begin{align}
frac{(x^4 -x^3+1)-(x^4+15x^2-5)}{left(sqrt{x^4 -x^3+1}+sqrt{x^4+15x^2-5},right)}
end{align}
Operate and calculate the limit.
edited Nov 22 '18 at 6:03
b00n heT
10.2k12234
answered Nov 22 '18 at 4:48
Alberto Torrejon ValenzuelaAlberto Torrejon Valenzuela
276
You should multiply by the conjugate of the expression, then proceed.
begin{align}
frac{left(sqrt{x^4 -x^3+1}-sqrt{x^4+15x^2-5},right)left(sqrt{x^4 -x^3+1}+sqrt{x^4+15x^2-5},right)}{left(sqrt{x^4 -x^3+1}+sqrt{x^4+15x^2-5},right)}\
end{align}
begin{align}
frac{(x^4 -x^3+1)-(x^4+15x^2-5)}{left(sqrt{x^4 -x^3+1}+sqrt{x^4+15x^2-5},right)}
end{align}
Operate and calculate the limit.
edited Nov 22 '18 at 6:03
b00n heT
10.2k12234
answered Nov 22 '18 at 4:48
Alberto Torrejon ValenzuelaAlberto Torrejon Valenzuela
276
edited Nov 22 '18 at 6:03
b00n heT
10.2k12234
edited Nov 22 '18 at 6:03
b00n heT
10.2k12234
edited Nov 22 '18 at 6:03
b00n heT
10.2k12234
10.2k12234
answered Nov 22 '18 at 4:48
Alberto Torrejon ValenzuelaAlberto Torrejon Valenzuela
276
answered Nov 22 '18 at 4:48
Alberto Torrejon ValenzuelaAlberto Torrejon Valenzuela
276
answered Nov 22 '18 at 4:48
Alberto Torrejon ValenzuelaAlberto Torrejon Valenzuela
276
276
negative! that's cool!
– Christopher Marley
Nov 22 '18 at 4:52
What does that mean? Sorry I`m not a fully native english speaker so those two words in the same sentence make me confuse.
– Alberto Torrejon Valenzuela
Nov 22 '18 at 4:54
and then, should i evaluate de sign of de function to see the lateral limits?
– Franco Cabrera
Nov 22 '18 at 4:54
No need if you just want to give a result of the limit. If you want to study the function then I suppose you should
– Alberto Torrejon Valenzuela
Nov 22 '18 at 4:56
Alberto, despues de eso tendria que evaluar el signo de f? ver si los limites laterales son iguales y obtener su valor?
– Franco Cabrera
Nov 22 '18 at 4:56
|
show 3 more comments
negative! that's cool!
– Christopher Marley
Nov 22 '18 at 4:52
What does that mean? Sorry I`m not a fully native english speaker so those two words in the same sentence make me confuse.
– Alberto Torrejon Valenzuela
Nov 22 '18 at 4:54
and then, should i evaluate de sign of de function to see the lateral limits?
– Franco Cabrera
Nov 22 '18 at 4:54
No need if you just want to give a result of the limit. If you want to study the function then I suppose you should
– Alberto Torrejon Valenzuela
Nov 22 '18 at 4:56
Alberto, despues de eso tendria que evaluar el signo de f? ver si los limites laterales son iguales y obtener su valor?
– Franco Cabrera
Nov 22 '18 at 4:56
negative! that's cool!
– Christopher Marley
Nov 22 '18 at 4:52
negative! that's cool!
– Christopher Marley
Nov 22 '18 at 4:52
What does that mean? Sorry I`m not a fully native english speaker so those two words in the same sentence make me confuse.
– Alberto Torrejon Valenzuela
Nov 22 '18 at 4:54
What does that mean? Sorry I`m not a fully native english speaker so those two words in the same sentence make me confuse.
– Alberto Torrejon Valenzuela
Nov 22 '18 at 4:54
and then, should i evaluate de sign of de function to see the lateral limits?
– Franco Cabrera
Nov 22 '18 at 4:54
and then, should i evaluate de sign of de function to see the lateral limits?
– Franco Cabrera
Nov 22 '18 at 4:54
No need if you just want to give a result of the limit. If you want to study the function then I suppose you should
– Alberto Torrejon Valenzuela
Nov 22 '18 at 4:56
No need if you just want to give a result of the limit. If you want to study the function then I suppose you should
– Alberto Torrejon Valenzuela
Nov 22 '18 at 4:56
Alberto, despues de eso tendria que evaluar el signo de f? ver si los limites laterales son iguales y obtener su valor?
– Franco Cabrera
Nov 22 '18 at 4:56
Alberto, despues de eso tendria que evaluar el signo de f? ver si los limites laterales son iguales y obtener su valor?
– Franco Cabrera
Nov 22 '18 at 4:56
|
show 3 more comments
Another manipulation:
Let $a,b >0$, real.
$a-b=(√a-√b)(√a+√b)$, then
$√a-√b = dfrac{a-b}{√a+√b}.$
$a: = (x^4-x^3+1)^{1/2},$
$b=(x^4+15x^2-5)^{1/2}.$
$small{f(x):=}$
$small {=dfrac{-x^3-15x^2+6}{(x^4-x^3+1)^{1/2}+(x^4+15x^2-5)^{1/2}}}$
Note : For large x (positive) the numerator is negative, the denominator positive: Increasing the denominator makes the fraction bigger.
Hence
$f(x) lt $
$dfrac{-x^3+6}{(x^4+1)^{1/2}+(x^4+15x^2)^{1/2}} lt $
$dfrac{-x^3+6}{√2x^2+√2x^2}=$
$dfrac{-x^3+6}{(2√2)x^2}.$
Take the limit.
answered Nov 22 '18 at 9:34
Peter SzilasPeter Szilas
10.8k2720
add a comment |
Another manipulation:
Let $a,b >0$, real.
$a-b=(√a-√b)(√a+√b)$, then
$√a-√b = dfrac{a-b}{√a+√b}.$
$a: = (x^4-x^3+1)^{1/2},$
$b=(x^4+15x^2-5)^{1/2}.$
$small{f(x):=}$
$small {=dfrac{-x^3-15x^2+6}{(x^4-x^3+1)^{1/2}+(x^4+15x^2-5)^{1/2}}}$
Note : For large x (positive) the numerator is negative, the denominator positive: Increasing the denominator makes the fraction bigger.
Hence
$f(x) lt $
$dfrac{-x^3+6}{(x^4+1)^{1/2}+(x^4+15x^2)^{1/2}} lt $
$dfrac{-x^3+6}{√2x^2+√2x^2}=$
$dfrac{-x^3+6}{(2√2)x^2}.$
Take the limit.
answered Nov 22 '18 at 9:34
Peter SzilasPeter Szilas
10.8k2720
add a comment |
Another manipulation:
Let $a,b >0$, real.
$a-b=(√a-√b)(√a+√b)$, then
$√a-√b = dfrac{a-b}{√a+√b}.$
$a: = (x^4-x^3+1)^{1/2},$
$b=(x^4+15x^2-5)^{1/2}.$
$small{f(x):=}$
$small {=dfrac{-x^3-15x^2+6}{(x^4-x^3+1)^{1/2}+(x^4+15x^2-5)^{1/2}}}$
Note : For large x (positive) the numerator is negative, the denominator positive: Increasing the denominator makes the fraction bigger.
Hence
$f(x) lt $
$dfrac{-x^3+6}{(x^4+1)^{1/2}+(x^4+15x^2)^{1/2}} lt $
$dfrac{-x^3+6}{√2x^2+√2x^2}=$
$dfrac{-x^3+6}{(2√2)x^2}.$
Take the limit.
answered Nov 22 '18 at 9:34
Peter SzilasPeter Szilas
10.8k2720
Another manipulation:
Let $a,b >0$, real.
$a-b=(√a-√b)(√a+√b)$, then
$√a-√b = dfrac{a-b}{√a+√b}.$
$a: = (x^4-x^3+1)^{1/2},$
$b=(x^4+15x^2-5)^{1/2}.$
$small{f(x):=}$
$small {=dfrac{-x^3-15x^2+6}{(x^4-x^3+1)^{1/2}+(x^4+15x^2-5)^{1/2}}}$
Note : For large x (positive) the numerator is negative, the denominator positive: Increasing the denominator makes the fraction bigger.
Hence
$f(x) lt $
$dfrac{-x^3+6}{(x^4+1)^{1/2}+(x^4+15x^2)^{1/2}} lt $
$dfrac{-x^3+6}{√2x^2+√2x^2}=$
$dfrac{-x^3+6}{(2√2)x^2}.$
Take the limit.
answered Nov 22 '18 at 9:34
Peter SzilasPeter Szilas
10.8k2720
edited Nov 22 '18 at 9:44
answered Nov 22 '18 at 9:34
Peter SzilasPeter Szilas
10.8k2720
answered Nov 22 '18 at 9:34
Peter SzilasPeter Szilas
10.8k2720
answered Nov 22 '18 at 9:34
Peter SzilasPeter Szilas
10.8k2720
10.8k2720
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Try multiplying and dividing by the conjugate root to eliminate the roots.
– D.B.
Nov 22 '18 at 4:37
thats what i did, but i got stuck
– Franco Cabrera
Nov 22 '18 at 4:41
math.stackexchange.com/questions/2959619/…
– lab bhattacharjee
Nov 22 '18 at 4:41
Remember that intuitively, when you work with limits of polynomials, you can often just replace the polynomial with it's leading term and get the right answer. In this case you get $0$
– Ovi
Nov 22 '18 at 4:49