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TG EAMCET 2 May 2025 Shift 2 Paper

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Question : 61 of 160

Marks: +1, -0

If

f(x)=\frac{x (a^x-1)}{1-\cos x}

g(x)=\frac{x (1-a^x)}{a^x (\sqrt{1-x^2}-\sqrt{1+x^2})}

\lim\limits_{x\to 0}(f(x)-g(x))=

$3\log a$
$e^a$
$2\log a$
$\log a$

Solution:

\text{ Given, } f(x)=\frac{x (a^x-1)}{1-\cos x}

\text{ And } g(x)=\frac{x (1-a^x)}{a^x (\sqrt{1-x^2}-\sqrt{1+x^2})}

\therefore \lim\limits_{x\to 0}f(x)=\lim\limits_{x\to 0}\frac{x (a^x-1)}{1-\cos x}

=\lim\limits_{x\to 0}\frac{x \frac{a^x-1}{x}}{\frac{1-\cos x}{x}}

=\lim\limits_{x\to 0}\frac{a^x-1}{x}\times\frac{x^2}{1-\cos x}

=\lim\limits_{x\to 0}\frac{a^x-1}{x}\times\lim\limits_{x\to 0}\frac{x^2}{1+\cos x}

=\log_a a\times 22\log_a a\lim\limits_{x\to 0}g(x)

=\lim\limits_{x\to 0}\frac{x (1-a^x)}{a^x (\sqrt{1-x^2}-\sqrt{1+x^2})}

=\lim\limits_{x\to 0}\frac{x (1-a^x) (\sqrt{1-x^2}+\sqrt{1+x^2})}{a^x (\sqrt{1-x^2}-\sqrt{1+x^2})}

\; (\sqrt{1-x^2}+\sqrt{1+x^2})

=\lim\limits_{x\to 0}\frac{x (1-a^x) (\sqrt{1-x^2}+\sqrt{1-x^2})}{a^x (1-x^2-1+x^2)}

=\lim\limits_{x\to 0}\frac{x (1-a^x) (\sqrt{1-x^2}+\sqrt{1-x^2})}{a^x (-2 x^2)}

=\lim\limits_{x\to 0}\frac{(a^x-1) (\sqrt{1-x^2}+\sqrt{1-x^2})}{a^x (2 x)}

=\lim\limits_{x\to 0}\frac{a^x-1}{x}\times\lim\limits_{x\to 0}\frac{\sqrt{1-x^2}+\sqrt{1-x^2}}{2 a^x}

=\log_a a\times\frac{1+1}{2(1)}

=\log_a a(1)=\log a

\therefore \lim\limits_{x\to 0}(f(x)-g(x))

=\lim\limits_{x\to 0}f(x)-\lim\limits_{x\to 0}g(x)

=2\log_a a-\log a=\log a

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