Question

question:-
lim x =>0 (ab)^x - a^x - b^x +1 / x^2​

note:-
(do not use L hospital law)
use trigonometry formulas ​

Answer 1

ANSWER:

$\displaystyle \lim_{x \to 0} \sf \frac{ (ab)^x - a^x - b^x +1}{ x^2}=2- \left(\dfrac{1}{ (sin^{-1}a)^2} \right)$

(or)

$\displaystyle \lim_{x \to 0} \sf \frac{ (ab)^x - a^x - b^x +1}{ x^2} = 2- \left(\dfrac{1}{ (cos^{-1}b)^2} \right)$

GIVEN:

$\displaystyle \lim_{x \to 0} \sf \frac{ (ab)^x - a^x - b^x +1}{ x^2}$

TO FIND:

$\displaystyle \lim_{x \to 0} \sf \frac{ (ab)^x - a^x - b^x +1}{ x^2}$

Without using L' Hospital's rule.

EXPLANATION:

$\displaystyle \lim_{x \to 0} \sf \frac{ (ab)^x - a^x - b^x +1}{ x^2}$

$\displaystyle \lim_{x \to 0} \sf \frac{ a^xb^x - a^x - b^x +1}{ x^2}$

$\displaystyle \lim_{x \to 0} \sf \frac{ a^x(b^x -1)- 1(b^x - 1)}{ x^2}$

$\displaystyle \lim_{x \to 0} \sf \frac{( a^x - 1)(b^x -1)}{ x^2}$

Substitute a = sin x and b = cos x

$\displaystyle \lim_{x \to 0} \sf \frac{( (sinx)^x - 1)((cosx)^x -1)}{ x^2}$

$\displaystyle \lim_{x \to 0} \sf \frac{\left( (sinx)^x \times \dfrac{ {x}^{x} }{ {x}^{x} } - 1\right)\left((cosx)^x \times \dfrac{ {x}^{x} }{ {x}^{x} }-1\right)}{ x^2}$

Sin x / x = cos x / x = 1

$\displaystyle \lim_{x \to 0} \sf \frac{( {x}^{x} - 1)( {x}^{x} -1)}{ x^2}$

(A - B)(A - B) = (A - B)²

$\displaystyle \lim_{x \to 0} \sf \frac{( {x}^{x} - 1)^{2} }{ x^2}$

(A - B)² = A² - 2AB + B²

$\displaystyle \lim_{x \to 0} \sf \frac{ {x}^{2x} - 2 {x}^{x} + 1}{ x^2}$

1 = sin² x + cos² x

$\displaystyle \lim_{x \to 0} \sf \frac{ {x}^{2x} - 2 {x}^{x} + {sin}^{2}x + {cos}^{2}x }{ x^2}$

$\bold{\dfrac{ {x}^{n} }{ {x }^{n} } = {x}^{m - n} }$

$\displaystyle \lim_{x \to 0} \sf \frac{ {x}^{2x}}{ x^2} - \dfrac{ 2 {x}^{x}}{ x^2} + \dfrac{{sin}^{2}x}{ x^2} + \dfrac{ {cos}^{2}x }{ x^2}$

$\displaystyle \lim_{x \to 0} \sf x^x\left(\frac{ {x}^{x}-2}{ x^2} \right) + \dfrac{{sin}^{2}x}{ x^2} + \dfrac{ {cos}^{2}x }{ x^2}$

Sin² x / x² = cos² x / x² = 1

$\displaystyle \lim_{x \to 0} \sf \ \ \ x^x\left(\frac{ {x}^{x}-2}{ x^2} \right)+ 1 + 1$

$\sf a = {sin} \ x$

$\sf x = {sin}^{ - 1} a$

$\sf x^2 = (sin^{-1}a)^2$

$\sf 0^0 \left(\dfrac{ 0^0-2}{ (sin^{-1}a)^2} \right) + 1 + 1$

$\bold{\large{0^0 = 1}}$

$\sf \left(\dfrac{ -1}{ (sin^{-1}a)^2} \right) + 2$

$\sf 2- \left(\dfrac{1}{ (sin^{-1}a)^2} \right)$

$\displaystyle \lim_{x \to 0} \sf \frac{ (ab)^x - a^x - b^x +1}{ x^2} = 2- \left(\dfrac{1}{ (sin^{-1}a)^2} \right)$

Similarly we can use

$\sf b = {cos} \ x$

$\sf x = {cos}^{ - 1} b$

$\sf x^2 = (cos^{-1}b)^2$

$\displaystyle \lim_{x \to 0} \sf \frac{ (ab)^x - a^x - b^x +1}{ x^2} = 2- \left(\dfrac{1}{ (cos^{-1}b)^2} \right)$