Math, asked by lalitsarda5693, 1 year ago

Limit x tends to 1 logx upon x-1

Answers

Answered by Anonymous
1

Answer:

1

Step-by-step explanation:

Method 1 --- L'Hospital's Rule

As x-->1, the expression ( log x ) / ( x - 1 ) tends to the form 0 / 0.  L'Hospital's rule applies.

The derivative of log x is 1/x.  The derivative of x-1 is 1.  Replacing numerator and denominator by the derivatives (L'Hospital's rule), we have:

\displaystyle\lim_{x\rightarrow 1}\frac{\ln x}{x-1}=\lim_{x\rightarrow1}\frac{1/x}{1}=\frac{1/1}{1} = 1

Method 2 --- Taylor series

Let u = 1 - x.  Then x = 1 - u and x --> 1 becomes u --> 0.  So we need the limit, as u --> 0, of  ln ( 1 - u ) / ( -u )  = - ln ( 1 - u ) / u.

Now recall the Taylor series

- ln ( 1 - u ) = u + u²/2 + u³/3 +...

Dividing by u, we have

- ln ( 1 - u ) / u = 1 + u/2 + u²/3 +...

Letting u --> 0, this tends to 1.

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