
How do you find the limit of $\left( \dfrac{1}{x}-\dfrac{1}{\left| x \right|} \right)$ where the $x$ approaches $0$ ?
Answer
571.5k+ views
Hint: From the question given it had been asked to find the limits of $\left( \dfrac{1}{x}-\dfrac{1}{\left| x \right|} \right)$ where the $x$ approaches $0$ . We can find the limits for a given function by using some basic formulae of limits. Let us assume the given function as $f\left( x \right)$.
Complete step by step answer:
Considering from the question we need to find the limit of $\left( \dfrac{1}{x}-\dfrac{1}{\left| x \right|} \right)$ when $x$ approaches to $0$ . Let us assume the given functions as $f\left( x \right)$ .
Therefore, $f\left( x \right)=\left( \dfrac{1}{x}-\dfrac{1}{\left| x \right|} \right)$
First of all, to find the limits of the given function, we have to verify whether the given function has limits or limits do not exist.
Knowing that, for the function $f\left( x \right)=\left( \dfrac{1}{x}-\dfrac{1}{\left| x \right|} \right)$ , the condition for the existence of limits is,
$\displaystyle \lim_{x \to 0}\text{ f}\left( x \right)$ exists , if $\displaystyle \lim_{x \to {{0}^{-}}}\text{ f}\left( x \right)\text{ = }\displaystyle \lim_{x \to {{0}^{+}}}\text{ f}\left( x \right)$
If the given function satisfies the above condition then only the limit will exist for the given function.
Now, further process is to verify the above written condition,
First of all, let us check the condition, $\displaystyle \lim_{x \to {{0}^{+}}}\text{ f}\left( x \right)$
Let us note that, as $x \to {{0}^{+}}\text{ , x 0}$
Therefore $\left| x \right|=x$
Now, by substituting in the given function we get,
$f\left( x \right)=\left( \dfrac{1}{x}-\dfrac{1}{\left| x \right|} \right)$
$\Rightarrow f\left( x \right)=\left( \dfrac{1}{x}-\dfrac{1}{x} \right)$
$\Rightarrow f\left( x \right)=0$
Now, let us check the another condition, $\displaystyle \lim_{x \to {{0}^{-}}}\text{ f}\left( x \right)\text{ }$
Let us know that, as $x \to {{0}^{-}}\text{ , x 0}$
Therefore $\left| x \right|=-x$
Now, by substituting in the given function we get,
$f\left( x \right)=\left( \dfrac{1}{x}-\dfrac{1}{\left| x \right|} \right)$
$\Rightarrow f\left( x \right)=\left( \dfrac{1}{x}-\left( -\dfrac{1}{-x} \right) \right)$
$\Rightarrow f\left( x \right)=\dfrac{2}{x}$
We can clearly observe that $\displaystyle \lim_{x \to {{0}^{-}}}\text{ f}\left( x \right)\text{ }\ne \text{ }\displaystyle \lim_{x \to {{0}^{+}}}\text{ f}\left( x \right)$
So, we can conclude that $\displaystyle \lim_{x \to 0}\text{ f}\left( x \right)\text{ }$ does not exist for the given function.
Note:
We should be well aware of the limits and their properties. We should be well aware of the condition for the limited existence. If for example we had forgotten and made a mistake and directly substituted zero in place of $x$ in $f\left( x \right)$ then we will write the limit as infinity which is the wrong conclusion so we should check for the existence of limits.
Complete step by step answer:
Considering from the question we need to find the limit of $\left( \dfrac{1}{x}-\dfrac{1}{\left| x \right|} \right)$ when $x$ approaches to $0$ . Let us assume the given functions as $f\left( x \right)$ .
Therefore, $f\left( x \right)=\left( \dfrac{1}{x}-\dfrac{1}{\left| x \right|} \right)$
First of all, to find the limits of the given function, we have to verify whether the given function has limits or limits do not exist.
Knowing that, for the function $f\left( x \right)=\left( \dfrac{1}{x}-\dfrac{1}{\left| x \right|} \right)$ , the condition for the existence of limits is,
$\displaystyle \lim_{x \to 0}\text{ f}\left( x \right)$ exists , if $\displaystyle \lim_{x \to {{0}^{-}}}\text{ f}\left( x \right)\text{ = }\displaystyle \lim_{x \to {{0}^{+}}}\text{ f}\left( x \right)$
If the given function satisfies the above condition then only the limit will exist for the given function.
Now, further process is to verify the above written condition,
First of all, let us check the condition, $\displaystyle \lim_{x \to {{0}^{+}}}\text{ f}\left( x \right)$
Let us note that, as $x \to {{0}^{+}}\text{ , x 0}$
Therefore $\left| x \right|=x$
Now, by substituting in the given function we get,
$f\left( x \right)=\left( \dfrac{1}{x}-\dfrac{1}{\left| x \right|} \right)$
$\Rightarrow f\left( x \right)=\left( \dfrac{1}{x}-\dfrac{1}{x} \right)$
$\Rightarrow f\left( x \right)=0$
Now, let us check the another condition, $\displaystyle \lim_{x \to {{0}^{-}}}\text{ f}\left( x \right)\text{ }$
Let us know that, as $x \to {{0}^{-}}\text{ , x 0}$
Therefore $\left| x \right|=-x$
Now, by substituting in the given function we get,
$f\left( x \right)=\left( \dfrac{1}{x}-\dfrac{1}{\left| x \right|} \right)$
$\Rightarrow f\left( x \right)=\left( \dfrac{1}{x}-\left( -\dfrac{1}{-x} \right) \right)$
$\Rightarrow f\left( x \right)=\dfrac{2}{x}$
We can clearly observe that $\displaystyle \lim_{x \to {{0}^{-}}}\text{ f}\left( x \right)\text{ }\ne \text{ }\displaystyle \lim_{x \to {{0}^{+}}}\text{ f}\left( x \right)$
So, we can conclude that $\displaystyle \lim_{x \to 0}\text{ f}\left( x \right)\text{ }$ does not exist for the given function.
Note:
We should be well aware of the limits and their properties. We should be well aware of the condition for the limited existence. If for example we had forgotten and made a mistake and directly substituted zero in place of $x$ in $f\left( x \right)$ then we will write the limit as infinity which is the wrong conclusion so we should check for the existence of limits.
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