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lim (1+logx-x)/(1-2x+x^2) = ? x->1
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\[\lim_{x \rightarrow 1} \frac{ 1+logx - x }{ 1-2x+x^2 }\]
now expand log x
log x, expansion? what is it ? can you please tell me
okk leave it
u knw l'hopital rule ??
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differentiate numerator and denominator seperatly
Yes I did that and got -1/2 as the answer. But it is not there in the options given.
\[\lim_{x \rightarrow 1} \frac{ 0+\frac{ 1 }{ x } -1}{ 0-2+2x }\]
its a silly Q but u type the Q correctly??
it is log x or ln x
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Its lnx
Please provide the final answer to it. I got it -1/2.
its coming -1/2 only
Ok, with the expansion too I am getting -1/2. So may be its the mistake of the options given for the question. Thank you very much.
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