Question:
The value of $f(0)$ so that $\frac{\left(-e^{x} +2^{x}\right)}{x}$may be continuous at $x = 0$ is
The value of $f(0)$ so that $\frac{\left(-e^{x} +2^{x}\right)}{x}$may be continuous at $x = 0$ is
Updated On: Apr 19, 2024
- $log\left(\frac{1}{2}\right)$
- 0
- 4
- - 1 + log 2
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The Correct Option is D
Solution and Explanation
$f \left(x\right)=\frac{-e^{x} + 2^{x}}{x}$
$=\frac{1}{x}\left[-\left(1+\frac{x}{1!}+\frac{x^{2}}{2!}+\frac{x^{3}}{3!}+..\right)+1+\frac{log 2}{1!}x+\frac{\left(bg 2\right)^{2}}{2!}x^{2}+\frac{\left(log 2\right)^{3}}{3!}x^{3}+..\right]$
$f \left(x\right)=log 2-1+\frac{x}{2!}\left\{\left(log 2\right)^{2} -\right\}
+\frac{x^{2}}{3!}\left\{\left(log^{2}\right)^{3}-1\right\}+....$
Putting $x = 0$, we get
$f\left(0\right) = log 2 - 1 + 0 + 0 + .... = - 1 + log \,2.$
$=\frac{1}{x}\left[-\left(1+\frac{x}{1!}+\frac{x^{2}}{2!}+\frac{x^{3}}{3!}+..\right)+1+\frac{log 2}{1!}x+\frac{\left(bg 2\right)^{2}}{2!}x^{2}+\frac{\left(log 2\right)^{3}}{3!}x^{3}+..\right]$
$f \left(x\right)=log 2-1+\frac{x}{2!}\left\{\left(log 2\right)^{2} -\right\}
+\frac{x^{2}}{3!}\left\{\left(log^{2}\right)^{3}-1\right\}+....$
Putting $x = 0$, we get
$f\left(0\right) = log 2 - 1 + 0 + 0 + .... = - 1 + log \,2.$
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Concepts Used:
Continuity
A function is said to be continuous at a point x = a, if
limx→a
f(x) Exists, and
limx→a
f(x) = f(a)
It implies that if the left hand limit (L.H.L), right hand limit (R.H.L) and the value of the function at x=a exists and these parameters are equal to each other, then the function f is said to be continuous at x=a.
If the function is undefined or does not exist, then we say that the function is discontinuous.
Conditions for continuity of a function: For any function to be continuous, it must meet the following conditions:
- The function f(x) specified at x = a, is continuous only if f(a) belongs to real number.
- The limit of the function as x approaches a, exists.
- The limit of the function as x approaches a, must be equal to the function value at x = a.