@ Sibgakhan

Q#01
Solve the two sided inequality and show the solution on real line
7 < 1-2x ≤ 10

Solution:
=> 7 < 1-2x ≤ 10
multiply through out by (-)
=> -10 ≤ 2x-1 <-7
=> -9 ≤ 2x < -7
Divide through out by 2
=> -9/2 ≤ x < -7/2
Hence the resultant solution is semi close semi open interval [-9/2 , -7/2) on real line.

Q#03
Simplify, then apply the rules of limit to evaluate
lim x->3 x3(power3)-5x2(power2)+6x/x2(power2)-9

Solution:
lim x->3 {x3(power3)-5x2(power2)+6x/x2(power2)-9} -------->(A) {c/0 form}
L' Hospital Rule is applied only if function has the form either {0/0} or {∞/∞}.
Now we proceed as follows
=> lim {x->3 x^3-5x^2+6x}/x^2 - 9
Nominator can be expanded as
x^3-5x^2+6x = x^3-3x^2-2x^2+6x = x^2*(x-3)-2x*(x-3) = (x^2-2x)*(x-3) ---->(i)
Also
x^2 - 9 = (x+3)*(x-3) ------>(ii)
Now using (i) and (ii) in (A)
=> lim x->3 {(x^2-2x)*(x-3)}/(x+3)*(x-3)
=> lim x->3 {(x^2-2x)}/(x+3)
Applying limit
=> {3^2 -2*3}/3+3
=> {9-6}/6
=> 3/6
=> 1/2 Ans

Best Regards!

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