Question

Evaluate
$\rm \displaystyle \lim_{n\to 3}\ \bigg[\frac{1}{x-3}-\frac{2}{x^{2}-4x+3}\bigg]$

Answer 1

See the image and ask if there is any problem.

Answer 2

Solution :

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i ) 1/(x-3) - 2/(x²-4x+3)

= 1/(x-3) - 2/[(x-1)(x-3)]

= 1/(x-3)[1 - 2/(x-1)]

= 1/(x-3)[ ( x-1-2)/(x-1) ]

= 1/(x-3)[(x-3)/(x-1)]

After cancellation, we get

= 1/(x-1)
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$\rm \displaystyle \lim_{x\to 3}\ \bigg[\frac{1}{x-3}-\frac{2}{x^{2}-4x+3}\bigg]$

= $\rm \displaystyle \lim_{x\to 3}\ \bigg[\frac{1}{x-1}}\bigg]$

= 1/(3-1)

= 1/2

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