11Y = Z = 1/(2√2)
oops sorry [2][2]
Y = lim x2tan(1/x)/(√8x2 + 7x + 1)
x--> - ∞
Z = lim x2tan(1/x)/(√8x2 + 7x + 1)
x--> ∞
Find Y and Z
-
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7 Answers
106yup Y=-Z because if u consider cancelling the √x2 then u have to take it as lxl so it will be Y=-Z
1357it should be Z=-Y=1/2√2
in Y put x=-y you will get something
1This was how i did for Y, pl. spot the mistake .......
let t = 1/x ----> 0
Y = (tant/t)/(√8 + 7p + 1)
Y = 1/2√2 ................. wer did i miss out the minus SIgn??
106Y = \lim_{x->infinity}x*[tan(1/x)/(1/x)]/\sqrt{8x^2+7x+1}
Y = \lim_{x->infinity}x/\sqrt{8x^2+7x+1} = \lim_{x-> infinity}x/\left|x \right|\sqrt{8+7/x+1/x^2}
if.x->-infinity.. then..x/\left|x \right| = -1..
so it is Y=-1/2√2 and so Z=1/2√2 as in Z x/lxl = 1
sry in Y .. it shud have been x--> -∞ and i have written ∞ .. sry