IIT JEE , AIEEE Forum - Mathematics , Differential Calculus

mbhushankamath

Evaluate the limits

1. Li m log x -1

x? 3 _______

x-e

(x tends to 3)

2. Lim 3ⁿ⁺¹+ 4

n?? _______

3ⁿ⁺² +4

(n tends to infinity)

Where (n+1) is the power of 3 in numerator &

(n+2) is the power of 3 in denominator.

3. Lim 2^x + 3^x

x?? ______

3^x + 1

(x tends to infinity)

Where x is the power of 2 and 3 &

also x is the power of 3 in denominator.

vinu

Hi mbhushankamath,
2). divide num & denom by 3^(n+2)
[where ^ means power raised to]
as n->infinity u get ans as 1/3. {'coz 4/3^(n+1) ->0}

3). divide num & denom by 3^x
as x->infinity, 1/3^x->0 ; (2/3)^x ->0 {as 2/3 < 1}
So, u get ans as 1.

1). num can be written as log(x/e)
just substitute x=3;
ans is ...[log (3/e)] / (3-e) .

mbhushankamath

sorry, there is a slight change in this question and thanks for the other two vinu.

Evaluate the limit

1. Li m log x -1

x? e _______

(x tends to e) x-e

nunoxic

1/e

simple !

log 1 = log e

lim log x - log e/x-e
=lim log (x/e)/x-e
= lim log (e+h/e)/h
=lim log (1 + h/e)/h

use (1+x)^1/x = e

= lim h/e * 1/h
=1/e

THERE IS A SHORTCUT.......PROMISE THAT U WILL NOT KEEP IT CONCEALED !! NEVER !! GOD HELPS WHO HELPS OTHERS !!
HERE IT IS !!
Lim [ log x - log a ] / [ x - a ]
x --> a

is always 1/a

nunoxic

or even simpler.....

lim log x - 1 / x - e

Hospital

lim d.log x/dx
=lim 1/x
=1/e

Moderator

Dear mbhushankamath ,

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puneet

hiiiii

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