Prove that 1) lim [log (x-3)] \\ [x+2] =1 x-> -2 2) lim [logx -1] \\ [x - e] = 1/e x-> e

1
by Arunav5

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

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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

sir check the new question menu i have eddited this one