The value of $f$ at $x = 0$ so that function $f(x) = \frac{2^{x} - 2^{- x}}{x},\ x
eq 0$ is continuous at $x = 0$, is

Question

The value of $f$ at $x = 0$ so that function $f(x) = \frac{2^{x} - 2^{- x}}{x},\ x 
eq 0$ is continuous at $x = 0$, is

Dr. Abubacker · Accepted Answer

(d)
$\lim_{x ightarrow 0}\frac{2^{x} - 2^{- x}}{x} = \lim_{x ightarrow 0}\ 2^{x}\log{2 + 2^{- x}\log 2}$
[by L’ Hospital’s rule]
$= \log 4$
Since, the function is continuous at $x = 0$
$	herefore\ \ \ \ \ \ f(0) = \lim_{x ightarrow 0}\ f(x)\ \  \Rightarrow \ \ \ \ f(0) = \log 4$

Q. The value of $f$ at $x = 0$ so that function $f(x) = \frac{2^{x} - 2^{- x}}{x},\ x \neq 0$ is continuous at $x = 0$, is

Explanation

Reviewed by Dr. Abubacker

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