Question:
Let f (x) = $ \frac {1} {\sqrt {18-x^2}}$, ten value o f
$ \lim_{x \to 3} (\frac {f(x)-f(3)} {x-3}) $
Let f (x) = $ \frac {1} {\sqrt {18-x^2}}$, ten value o f
$ \lim_{x \to 3} (\frac {f(x)-f(3)} {x-3}) $
Updated On: May 5, 2024
- 0
- $-\frac {1} {9}$
- $-\frac {1} {3}$
- $\frac {1} {9}$
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The Correct Option is D
Solution and Explanation
$\lim _{x\to 3} \frac{\left(18 -x^{2}\right)^{-\frac{1}{2} }- \left(\frac{1}{3}\right)}{x-3} =\lim _{x\to 3} \frac{3 -\sqrt{18 -x^{2}}}{3\left(x-3\right) \sqrt{18 -x^{2}} } $
= $\lim _{x\to 3} \left[\frac{9-\left(18 -x^{2}\right)}{3\left(x-3\right) \sqrt{18 -x^{2}}\left(3+\sqrt{18 -x^{2}}\right)}\right]$
= $\lim _{x\to 3} \frac{x^{ 2} -9}{3\left(x-3\right) \sqrt{18 -x^{2} } \left[3+\sqrt{18 -x^{2}}\right]}$
= $\lim _{x\to 3} \frac{x+3}{3\sqrt{18-x^{2}} \left[3+\sqrt{18 -x^{2}}\right] }$
= $\frac{6}{3\left(3\right)\left(3+3\right)} =\frac{1}{9}$
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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).