# find the limit,if existsf(x)=ln(x-2)/(x-3), x->3

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### 2 Answers

We need to find lim x--> 3 [ ln(x - 2)/(x - 3)]

lim x--> 3 [ ln(x - 2)/(x - 3)]

substituting x = 3 gives us the indeterminate form 0/0, so we can use l'Hopital's rule and substitute the numerator and denominator by their derivatives.

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

substitute x = 3

=> 1

**The required value of the limit is 1.**

First, we'll substitute x by the value of accumulation point.

lim ln(x-2)/(x-3) = ln(3-2)/(3-3) = ln1/0 = 0/0

We've get an indetermination, so, we'll apply L'Hospital rule:

lim ln(x-2)/(x-3) = lim d[ln(x-2)]/dx/d(x-3)/dx

lim d[ln(x-2)]/dx/d(x-3)/dx = lim 1/(x-2)/1

We'll substitute again x by 2 and we'll get:

lim 1/(x-2)/1 = 1/(3-2) = 1** **

**The limit of the function f(x), if x approaches to 3, is lim ln(x-2)/(x-3) = 1**