The right hand and left hand limit of the function are respectively
- 1 and 1
- 1 and -1
- -1 and -1
- -1 and 1
The Correct Option is B
Solution and Explanation
$=\frac{e^{-\infty}-1}{e^{-\infty}+1}=\frac{0-1}{0+1}=-1$
Dividing both numerator and denominator by $e^{1 / x}$, we get
$\displaystyle\lim _{x \rightarrow 0}\left(\frac{1-e^{-1 / x}}{1+e^{-1 / x}}\right)$
$RHL =\displaystyle\lim _{x \rightarrow 0^{+}}\left(\frac{1-e^{-1 / x}}{1+e^{-1 / x}}\right)$
$=\frac{1-e^{-\infty}}{1+e^{-\infty}}=\frac{1-0}{1+0}=1$
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Concepts Used:
Limits
A function's limit is a number that a function reaches when its independent variable comes to a certain value. The value (say a) to which the function f(x) approaches casually as the independent variable x approaches casually a given value "A" denoted as f(x) = A.
If limx→a- f(x) is the expected value of f when x = a, given the values of ‘f’ near x to the left of ‘a’. This value is also called the left-hand limit of ‘f’ at a.
If limx→a+ f(x) is the expected value of f when x = a, given the values of ‘f’ near x to the right of ‘a’. This value is also called the right-hand limit of f(x) at a.
If the right-hand and left-hand limits concur, then it is referred to as a common value as the limit of f(x) at x = a and denote it by lim x→a f(x).