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Find the limit, if it exists. (Ifan answer does not exist, enter DNE.)x2 4xlimх+

Question
Find the limit, if it exists. (If
an answer does not exist, enter DNE.)
x2 4x
lim
х+
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Find the limit, if it exists. (If an answer does not exist, enter DNE.) x2 4x lim х+

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check_circleAnswer
Step 1

Given,

lim x x2 + 4x
х--со
Putting
х— —х then x —> со
lim — x + Vx2 -4х
хэоо
-х-Vx2-4х
lim x x2
— 4х) X
-х-ух°—4х
хэоо
2
(-х)2- (Vx-4х)
lim
(а? — b? — (а + b)(а — b))
—х-Vx2-4х
хноо
lim x*-(x?-4х)
-4х
хоо —х-ух*—
4х
= lim
.2
хноо —х-Vx-4х
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lim x x2 + 4x х--со Putting х— —х then x —> со lim — x + Vx2 -4х хэоо -х-Vx2-4х lim x x2 — 4х) X -х-ух°—4х хэоо 2 (-х)2- (Vx-4х) lim (а? — b? — (а + b)(а — b)) —х-Vx2-4х хноо lim x*-(x?-4х) -4х хоо —х-ух*— 4х = lim .2 хноо —х-Vx-4х

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Step 2

Further calcu...

х
= 4 lim
хоо — х-Vx2-4х
х
= 4 lim
4
1
х-со
х
х
1
= 4 lim
4
1.
-1 -
1-.
1
= 4 x
4
1.
-1-,
1
= 4 X
-1-1
= 4 x
-4x
-2
=-2
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х = 4 lim хоо — х-Vx2-4х х = 4 lim 4 1 х-со х х 1 = 4 lim 4 1. -1 - 1-. 1 = 4 x 4 1. -1-, 1 = 4 X -1-1 = 4 x -4x -2 =-2

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Tagged in

Math

Calculus

Limits

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