Question

limit of e^sinax - e^sin bx /tan 2x​

Answer 1

Given :  $\lim_{x \to 0} \dfrac{e^{\sin ax} -e^{\sin bx} }{\tan 2x}$

To Find :  evaluate limit

Solution:

$\lim_{x \to 0} \dfrac{e^{\sin ax} -e^{\sin bx} }{\tan 2x}$

Substituting  x = 0  

sinx = 0  , tan x = 0 Hence

(e⁰ - e⁰) /0

= (1 - 1)/0

= 0/0

Apply L' Hospital rule

Differentiate numerator and denominator separately

$\lim_{x \to 0} \dfrac{a\cos ax .e^{\sin ax} -b \cos bx .e^{\sin bx} }{2\sec^2 2x}$

Substituting  x = 0  

cos 0 = 1  

sec 0 = 1

=  (a (1)e⁰  -b (1) e⁰ ) / (2 * 1²)

=  (a - b)/2

$\lim_{x \to 0} \dfrac{e^{\sin ax} -e^{\sin bx} }{\tan 2x}=\dfrac{a-b}{2}$

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