TIFR CSE 2011 | Part A | Question: 17

Question

What is $$\lim_{x \to 0} \frac{2^x-1}{x}$$

• A. $0$
• B. $\log_2(e)$
• C. $\log_e(2)$
• D. $1$
• E. None of the above

Answer 1

Since we have a $\frac{0}{0}$ form, we can apply the L'Hôpital's rule.

$$\begin{align}
L &= \lim_{x \to 0}\frac{2^x-1}{x}\\[1em]
&= \lim_{x \to 0}\frac{\frac{d}{dx}(2^x-1)}{\frac{d}{dx}x}\\[1em]
&= \lim_{x \to 0}\frac{2^x\ln 2}{1}\\[1em]
&= \ln 2\\[1em]
L &= \log_e{(2)}
\end{align}$$

Hence, option c is correct.

Comments

Derivative of a^x  is always a^x(logx to base e).....not base 2

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$\ln$ is by default in base $e.$

see here