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7.2 Q.I · Q21

Q.lim⁡z→2[z2−5z+6z2−4]\displaystyle\lim_{z\to 2}\left[\frac{z^2-5z+6}{z^2-4}\right]

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✓ Free question

z2−5z+6=(z−2)(z−3)z^2-5z+6=(z-2)(z-3) and z2−4=(z−2)(z+2)z^2-4=(z-2)(z+2). Cancelling the common factor (z−2)(z-2) (valid since z→2z\to2 but z≠2z\ne2): z−3z+2\dfrac{z-3}{z+2}. At z=2z=2: 2−32+2=−14\dfrac{2-3}{2+2}=-\dfrac14.

✓Final answer

−14-\dfrac{1}{4}

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