Answers

2015-03-09T18:17:57+05:30

This Is a Certified Answer

×
Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.
1)  This question does not seem to be correctly written. please check.  Is it x+3 in the numerator ?
         Lim     [ Log (x+3) ] / (x+2)
       x -> -2

   Let x + 2 = h    then  as  x ->2,  h -> 0.
 Hence  Lim h-> 0,  [ Log (h+1) ] / h
    
We use the Taylor series expansion for Log (1+h) = h - h²/2 + h³/3 - h⁴/4 + ...
Hence the limit is
   Lim h->0    h * [1 - h/2 + h²/3 - h³/4 + .... ] / h
  = Lim  h-> 0    [ 1 - h/2 + h²/3 - h³/4 +  ... ] 
  = 1
=======================
2)
 Let  x/e -1 = h .        Hence,  as  x -> e,  h -> 0.

Limit is Lim x->0    [ log x - 1 ] / [ x - e ]  = [ log x  - Log e ] / [ e * (x/e - 1) ]
      = Lim  x -> e      1/e *  Log x/e / [s/e - 1 ]
   =  Lim  h -> 0      1/e *  [ Log (1+h) ] / h

We use the Taylor's series for expansion of Log (1 + h)  as in the above problem.
   The limit is  Lim  h-> 0   1/e * [ h - h²/2 + h³ / 3 - h⁴/4 + .... ]  / h
       = 1/e *  Lim  h -> 0    1 - h/2 + h²/3 - h³/4  + ...
       = 1 / e *  1
       = 1/e

1 5 1
sir check the new question menu i have eddited this one