ₓ→ 0}{a^{x}-b^{x}}{e^{x}-1} is equal to :

Correct Answer is : log⁡(ab)≤ft(a/b)log(ba)

Correct Answer is : log⁡(ab)≤ft(a/b)log(ba)

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UPSC NDA Math Model Paper 7

Question : 59 of 120

Marks: +1, -0

\lim\limits_{x\to 0}\frac{a^{x}-b^{x}}{e^{x}-1}

is equal to :

Solution:

Here, we have to find the limit of

\lim\limits_{x\to 0}\frac{a^{x}-b^{x}}{e^{x}-1}

The expression

\frac{a^{x}-b^{x}}{e^{x}-1}

can be re-written as:

\Rightarrow\frac{a^{x}-b^{x}}{e^{x}-1}=\left[\frac{\frac{a^{x}-1}{x}-\frac{b^{x}-1}{x}}{\frac{e^{x}-1}{x}}\right]

Now by applying limits on both the sides of the above equation we get

\Rightarrow\lim\limits_{x\to 0}\frac{a^{x}-b^{x}}{e^{x}-1}

=\lim\limits_{x\to 0}\left[\frac{\frac{a^{x}-1}{x}-\frac{b^{x}-1}{x}}{\frac{e^{x}-1}{x}}\right]

As we know that,

\lim\limits_{x\to a}\left[\frac{f(x)}{g(x)}\right]=x\to a\lim\limits_{x\to a}g(x)

, provided

\lim\limits_{x\to a}g(x)\neq 0

\Rightarrow\lim\limits_{x\to 0}\left[\frac{\frac{a^{x}-1}{x}-\frac{b^{x}-1}{x}}{\frac{e^{x}-1}{x}}\right]

=\frac{\left[\lim\limits_{x\to 0}\frac{a^{x}-1}{x}\right]-\left[\lim\limits_{x\to 0}\frac{b^{x}-1}{x}\right]}{\left[\lim\limits_{x\to 0}\frac{e^{x}-1}{x}\right]}

As we know that,

\lim\limits_{x\to 0}\left[\frac{a^{x}-1}{x}\right]=\log a,\;a>0

and

\lim\limits_{x\to 0}\left[\frac{e^{x}-1}{x}\right]=1

\Rightarrow\lim\limits_{x\to 0}\frac{a^{x}-b^{x}}{e^{x}-1}=\log\frac{a}{b}

Hence, option

A

is true.

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