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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