Question

2. দেখাও যে,
ti) ) lim
=
= 8log.2
(
x1 -1​

Answer 1

Given :  $\lim_{x \to 1} \frac{4^x - 4 }{x - 1 } = 8 log_{e} 2$

To Find :  To Prove

Solution:

$\lim_{x \to 1} \frac{4^x - 4 }{x - 1 } = 8 log_{e} 2$

LHS =

$\lim_{x \to 1} \frac{4^x - 4 }{x - 1 }$

if we put x  = 1

then Numerator = 4¹ - 4 = 0

Denominator = 1 - 1 = 0

hence its in form of    0/0

Hence we can apply L'Hospital Rule

and differentiate numerator & Denominator

d(4ˣ  - 4)/dx  =  4ˣ  $log_{e} 4$

d(x - 1)/dx  =  1

=  $\lim_{x \to 1} \frac{4^x log_{e}4 }{ 1}$

putting x = 1

=   4¹  $log_{e} 4$

= 4 $log_{e} 4$

= 4 $log_{e} 2^2$

= 4 * 2 $log_{e} 2$

= 8  $log_{e} 2$

= RHS

LHS = RHS

QED

Hence Proved

$\lim_{x \to 1} \frac{4^x - 4 }{x - 1 } = 8 log_{e} 2$

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