Question

prove that

1) lim $\frac{log(x-3)}{x+2} = 1$
x-> -2

2) lim $\frac{logx-1}{x-e} = \frac{1}{e}$
x->e

Answer 1

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

$\lim_{x \to -2} \frac{Log\ (x+3)}{x+2}\\\\Let\ x+2=h,\ as\ x \to -2,\ \ h \to 0.\\\\= \lim_{h \to 0} \frac{Log\ (h+1)}{h}\\\\=Lim_{h \to 0} \frac{h-\frac{h^2}{2}+\frac{h^3}{3}-\frac{h^4}{4}+...}{h}\\\\=Lim_{h \to 0} [1 -\frac{h}{2}+\frac{h^2}{3}-\frac{h^3}{4}+...]\\\\=1$

We use the Taylor series expansion for Log (1+h) = h - h²/2 + h³/3 - h⁴/4 + ...

=======================
2)
 Let  x/e -1 = h .        Hence,  as  x -> e,  h -> 0.

$Limit\ is\ Lim_{x \to 0}\ \frac{log x - 1}{x-e}\\\\= \frac{log x - Log e}{e * (x/e - 1)}\\\\=Lim_{x \to e}\ \frac{1}{e}* \frac{Log\ \frac{x}{e}} {s/e - 1}\\\\= Lim_{h \to 0}\frac{1}{e}* \frac{ Log (1+h)} {h}$

We use the Taylor's series for expansion of Log (1 + h)  as in the above problem.

The limit is 
$Lim_{h \to 0}\ \frac{1}{e} * \frac{[ h - \frac{h^2}{2} + \frac{h^3}{3} - \frac{h^4}{4} + .... ]} {h}\\\\=\frac{1}{e} * Lim_{h \to 0}\ \ 1 - \frac{h}{2} + \frac{h^2}{3} - \frac{h^3}{4}+...\\\\= \frac{1}{e} *1\\\\=\frac{1}{e}$