Factor square root out of quotient
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How do I get from the first expression to the second?
The only reason the limits are included is because on WolframAlpha it mentioned that this was the case for large negative numbers of x:
$lim_{xto -infty} frac{sqrt{(x^2 + 2)}}{x} = lim_{xto -infty} -sqrt{frac{(x^2 + 2)}{x^2}}$
limits radicals factoring quotient-group
asked Nov 16 at 13:59
MierD. Luffy
51
add a comment |
up vote
0
down vote
favorite
How do I get from the first expression to the second?
The only reason the limits are included is because on WolframAlpha it mentioned that this was the case for large negative numbers of x:
$lim_{xto -infty} frac{sqrt{(x^2 + 2)}}{x} = lim_{xto -infty} -sqrt{frac{(x^2 + 2)}{x^2}}$
limits radicals factoring quotient-group
asked Nov 16 at 13:59
MierD. Luffy
51
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
How do I get from the first expression to the second?
The only reason the limits are included is because on WolframAlpha it mentioned that this was the case for large negative numbers of x:
$lim_{xto -infty} frac{sqrt{(x^2 + 2)}}{x} = lim_{xto -infty} -sqrt{frac{(x^2 + 2)}{x^2}}$
limits radicals factoring quotient-group
asked Nov 16 at 13:59
MierD. Luffy
51
How do I get from the first expression to the second?
The only reason the limits are included is because on WolframAlpha it mentioned that this was the case for large negative numbers of x:
$lim_{xto -infty} frac{sqrt{(x^2 + 2)}}{x} = lim_{xto -infty} -sqrt{frac{(x^2 + 2)}{x^2}}$
limits radicals factoring quotient-group
limits radicals factoring quotient-group
asked Nov 16 at 13:59
MierD. Luffy
51
asked Nov 16 at 13:59
MierD. Luffy
51
asked Nov 16 at 13:59
MierD. Luffy
51
asked Nov 16 at 13:59
MierD. Luffy
51
asked Nov 16 at 13:59
MierD. Luffy
51
51
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
up vote
0
down vote
accepted
Multiply the numerator and denominator by $-1$:
$$lim_{xto -infty} frac{sqrt{(x^2 + 2)}}{x} = lim_{xto -infty} frac{-sqrt{(x^2 + 2)}}{-x}=lim_{xto -infty} frac{-sqrt{(x^2 + 2)}}{|x|}=\
=lim_{xto -infty} frac{-sqrt{(x^2 + 2)}}{sqrt{x^2}}=lim_{xto -infty} -sqrt{frac{(x^2 + 2)}{x^2}}$$
answered Nov 16 at 16:56
farruhota
17.6k2736
Thank you. Its super clear now.
– MierD. Luffy
Nov 17 at 1:08
You are welcome. Good luck.
– farruhota
Nov 17 at 2:38
add a comment |
up vote
0
down vote
Notice that we have $$sqrt{x^2}=|x|$$
when $x<0$, we have $|x|=-x$
$$x=-|x|=-sqrt{x^2}$$
answered Nov 16 at 14:02
Siong Thye Goh
93.1k1462114
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
Multiply the numerator and denominator by $-1$:
$$lim_{xto -infty} frac{sqrt{(x^2 + 2)}}{x} = lim_{xto -infty} frac{-sqrt{(x^2 + 2)}}{-x}=lim_{xto -infty} frac{-sqrt{(x^2 + 2)}}{|x|}=\
=lim_{xto -infty} frac{-sqrt{(x^2 + 2)}}{sqrt{x^2}}=lim_{xto -infty} -sqrt{frac{(x^2 + 2)}{x^2}}$$
answered Nov 16 at 16:56
farruhota
17.6k2736
Thank you. Its super clear now.
– MierD. Luffy
Nov 17 at 1:08
You are welcome. Good luck.
– farruhota
Nov 17 at 2:38
add a comment |
up vote
0
down vote
accepted
Multiply the numerator and denominator by $-1$:
$$lim_{xto -infty} frac{sqrt{(x^2 + 2)}}{x} = lim_{xto -infty} frac{-sqrt{(x^2 + 2)}}{-x}=lim_{xto -infty} frac{-sqrt{(x^2 + 2)}}{|x|}=\
=lim_{xto -infty} frac{-sqrt{(x^2 + 2)}}{sqrt{x^2}}=lim_{xto -infty} -sqrt{frac{(x^2 + 2)}{x^2}}$$
answered Nov 16 at 16:56
farruhota
17.6k2736
Thank you. Its super clear now.
– MierD. Luffy
Nov 17 at 1:08
You are welcome. Good luck.
– farruhota
Nov 17 at 2:38
add a comment |
up vote
0
down vote
accepted
up vote
0
down vote
accepted
Multiply the numerator and denominator by $-1$:
$$lim_{xto -infty} frac{sqrt{(x^2 + 2)}}{x} = lim_{xto -infty} frac{-sqrt{(x^2 + 2)}}{-x}=lim_{xto -infty} frac{-sqrt{(x^2 + 2)}}{|x|}=\
=lim_{xto -infty} frac{-sqrt{(x^2 + 2)}}{sqrt{x^2}}=lim_{xto -infty} -sqrt{frac{(x^2 + 2)}{x^2}}$$
answered Nov 16 at 16:56
farruhota
17.6k2736
Multiply the numerator and denominator by $-1$:
$$lim_{xto -infty} frac{sqrt{(x^2 + 2)}}{x} = lim_{xto -infty} frac{-sqrt{(x^2 + 2)}}{-x}=lim_{xto -infty} frac{-sqrt{(x^2 + 2)}}{|x|}=\
=lim_{xto -infty} frac{-sqrt{(x^2 + 2)}}{sqrt{x^2}}=lim_{xto -infty} -sqrt{frac{(x^2 + 2)}{x^2}}$$
answered Nov 16 at 16:56
farruhota
17.6k2736
answered Nov 16 at 16:56
farruhota
17.6k2736
answered Nov 16 at 16:56
farruhota
17.6k2736
answered Nov 16 at 16:56
farruhota
17.6k2736
17.6k2736
Thank you. Its super clear now.
– MierD. Luffy
Nov 17 at 1:08
You are welcome. Good luck.
– farruhota
Nov 17 at 2:38
add a comment |
Thank you. Its super clear now.
– MierD. Luffy
Nov 17 at 1:08
You are welcome. Good luck.
– farruhota
Nov 17 at 2:38
Thank you. Its super clear now.
– MierD. Luffy
Nov 17 at 1:08
Thank you. Its super clear now.
– MierD. Luffy
Nov 17 at 1:08
You are welcome. Good luck.
– farruhota
Nov 17 at 2:38
You are welcome. Good luck.
– farruhota
Nov 17 at 2:38
add a comment |
up vote
0
down vote
Notice that we have $$sqrt{x^2}=|x|$$
when $x<0$, we have $|x|=-x$
$$x=-|x|=-sqrt{x^2}$$
answered Nov 16 at 14:02
Siong Thye Goh
93.1k1462114
add a comment |
up vote
0
down vote
Notice that we have $$sqrt{x^2}=|x|$$
when $x<0$, we have $|x|=-x$
$$x=-|x|=-sqrt{x^2}$$
answered Nov 16 at 14:02
Siong Thye Goh
93.1k1462114
add a comment |
up vote
0
down vote
up vote
0
down vote
Notice that we have $$sqrt{x^2}=|x|$$
when $x<0$, we have $|x|=-x$
$$x=-|x|=-sqrt{x^2}$$
answered Nov 16 at 14:02
Siong Thye Goh
93.1k1462114
Notice that we have $$sqrt{x^2}=|x|$$
when $x<0$, we have $|x|=-x$
$$x=-|x|=-sqrt{x^2}$$
answered Nov 16 at 14:02
Siong Thye Goh
93.1k1462114
answered Nov 16 at 14:02
Siong Thye Goh
93.1k1462114
answered Nov 16 at 14:02
Siong Thye Goh
93.1k1462114
answered Nov 16 at 14:02
Siong Thye Goh
93.1k1462114
93.1k1462114
add a comment |
add a comment |
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