Question:
The value of $\displaystyle\lim_{x \to 0} \frac {5^x-5^{-x}}{2x}=$ is :
The value of $\displaystyle\lim_{x \to 0} \frac {5^x-5^{-x}}{2x}=$ is :
Updated On: Apr 8, 2024
- 2 log 5
- 1
- 0
- log 5
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The Correct Option is D
Solution and Explanation
$\displaystyle \lim _{x \rightarrow 0} \frac{5^{x}-5^{-x}}{2 x}$
Applying L- Hospital's rule
$=\displaystyle\lim _{x \rightarrow 0} \frac{5^{x} \log 5+5^{-x} \log 5}{2} $
$=\frac{\log 5+\log 5}{2}=\log 5$
Applying L- Hospital's rule
$=\displaystyle\lim _{x \rightarrow 0} \frac{5^{x} \log 5+5^{-x} \log 5}{2} $
$=\frac{\log 5+\log 5}{2}=\log 5$
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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).