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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

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1) lim [log (x-3)] \\ [x+2] =1

x-> -2

2) lim [logx -1] \\ [x - e] = 1/e

x-> e

by Arunav5

Log in to add a comment

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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

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