Limits
3.0 Conditions for existence of Limit
3.0 Conditions for existence of Limit
$\mathop {\lim }\limits_{x \to a} f(x) $ exists if both $\mathop {\lim }\limits_{x \to {a^ - }} f(x)$, $\mathop {\lim }\limits_{x \to {a^ + }} f(x)$ exists and are equal.
i.e., Limit of a function exists if both left hand and right hand limits exists and are equal.
NOTE:
- If $\mathop {\lim }\limits_{x \to {a^ - }} f(x)$, $\mathop {\lim }\limits_{x \to {a^ + }} f(x)$ exists and $\mathop {\lim }\limits_{x \to {a^ - }} f(x)$ = $\mathop {\lim }\limits_{x \to {a^ + }} f(x)$ then $\mathop {\lim }\limits_{x \to a} f(x) $ = $\mathop {\lim }\limits_{x \to {a^ - }} f(x)$ = $\mathop {\lim }\limits_{x \to {a^ + }} f(x)$.
- The Limit of a function, if it exists is unique.
Question 7.
$f(x) = \left\{ \begin{gathered} 5x - 4{\text{ , x < 1}} \hspace{1em} \\ 4{x^3} - 3x{\text{ , x > 1}} \hspace{1em} \\ \end{gathered} \right.$
Find $\mathop {\lim }\limits_{x \to 1} f(x)$.
Solution:
First we find LHL of $f(x)$ at $x$=1:
$\mathop {\lim }\limits_{x \to {1^ - }} f(x) = \mathop {\lim }\limits_{x \to {1^ - }}( 5x - 4 )= 1$
next RHL of $f(x)$ at $x$=1:
$\mathop {\lim }\limits_{x \to {1^ + }} f(x) = \mathop {\lim }\limits_{x \to {1^ + }}( 4{x^3} - 3x) = 1$
LHL=RHL. So $\mathop {\lim }\limits_{x \to 1} f(x)$ = 1.
Note that even though $f(x)$ is not defined at $x$=1, Limit of $f(x)$ at $x$=1 exists finitely.
Question 8.
If $f(x) = \left\{ \begin{gathered} \frac{{\left| {x - 10} \right|}}{{x - 10}}{\text{ }},\;x \ne 10 \hspace{1em} \\ {\text{ }}0\;{\text{ }},\;x = 10 \hspace{1em} \\ \end{gathered} \right.$.
Find $\mathop {\lim }\limits_{x \to 10} f(x)$.
Solution:
We know that
$\left| r \right| = \left\{ \begin{gathered} r\ ,\ r > 0 \hspace{1em} \\ - r\ ,\ r < 0 \hspace{1em} \\ \end{gathered} \right.$
$ \Rightarrow f(x) = \left\{ \begin{gathered} \frac{{x - 10}}{{x - 10}}\ ,\ x - 10 > 0 \hspace{1em} \\ \frac{{ - (x - 10)}}{{x - 10}}\ ,\ x - 10 < 0 \hspace{1em} \\ 0\,\ x = 10 \hspace{1em} \\ \end{gathered} \right.$
$ \Rightarrow f(x) = \left\{ \begin{gathered} 1\ ,\ x > 10 \hspace{1em} \\ - 1\ ,\ x < 10 \hspace{1em} \\ 0\ ,\ x = 10 \hspace{1em} \\ \end{gathered} \right.$
RHL at $x=10$:
$\mathop {\lim }\limits_{x \to {{10}^ + }} f(x) = 1$
LHL at $x=10$:
$\mathop {\lim }\limits_{x \to {{10}^ - }} f(x) = - 1$
As RHL $ \ne $ LHL. $\mathop {\lim }\limits_{x \to 10} f(x)$ doesn't exist.
