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Evaluate following limits.
Solution : Let's do the first limit & in this case it sees like we will factor a z3 out of the numerator and denominator both. Remember that we just see at the denominator while determining the largest power of z here. In the numerator there is a larger power of z however we ignore it. We just see at the denominator while doing this! So doing the factoring gives,
While we take the limit we'll have to be a little careful. The first term in the numerator & denominator will both be zero. Though, the z3 in the numerator will be going to plus infinity in the limit and therefore the limit is,
The final limit is negative since we have a quotient of positive quantity & a negative quantity.
Now, let's take a look at the second limit. Note as well that the only different in the work is at the final "evaluation" step and therefore we'll pick up the work there.
In this case the z3 in the numerator gives -ve infinity in the limit as we are going to minus infinity & the power is odd. The answer is positive as we have a quotient of two negative numbers.
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Solve : 4x2+2x+3=0 Ans) x^2 + (1/2)x = -(3/4) (x+1/4)^2 = 1/16 - 3/4 = -11/16 implies x = (-1+i(11)^(1/2))/4 and its conjugate.
where does goes take place?
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