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# If$f(x)=\left\{ \begin{matrix} \sqrt{x-4}\text{ if }x>4 \\ 8-2x\text{ if }x<4 \\\end{matrix} \right.$determine whether$\underset{x\to 4}{\mathop{\lim }}\,f(x)$ exists.  Verified
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Hint: For finding out whether the limit exists, then we should find the left hand limit and right hand limit. If they are equal then the limit exists.

If $\underset{x\to 4}{\mathop{\lim }}\,f(x)\text{ }$exists then its left hand limit must be equal to its right hand limit, that is,
$\underset{x\to {{4}^{-}}}{\mathop{\lim }}\,f(x)=\underset{x\to {{4}^{+}}}{\mathop{\lim }}\,f(x)$
Now we will find the left hand limit of the given function, we get
$\underset{x\to {{4}^{-}}}{\mathop{\lim }}\,f(x)=\underset{x\to {{4}^{-}}}{\mathop{\lim }}\,(8-2x)$
We also know the limit of difference is the difference of the limits. So, the limit of difference of two functions is equal to the difference of individual limits of the functions, that is,
$\underset{x\to {{4}^{-}}}{\mathop{\lim }}\,f(x)=\underset{x\to {{4}^{-}}}{\mathop{\lim }}\,(8)-\underset{x\to {{4}^{+}}}{\mathop{\lim }}\,(2x)$
Now we know limit of a constant is always the constant, so the above equation can be written as,
$\underset{x\to {{4}^{-}}}{\mathop{\lim }}\,f(x)=8-\underset{x\to {{4}^{+}}}{\mathop{\lim }}\,(2x)$
Now applying the limits, we get
$\underset{x\to {{4}^{-}}}{\mathop{\lim }}\,f(x)=8-(2\times 4)=8-8$
$\underset{x\to {{4}^{-}}}{\mathop{\lim }}\,f(x)=0........(i)$
So the left hand limit exists and is equal to zero.
Now we will find the right hand limit of the given function, we get
$\underset{x\to {{4}^{+}}}{\mathop{\lim }}\,f(x)=\underset{x\to {{4}^{+}}}{\mathop{\lim }}\,(\sqrt{x-4})$
Now applying the limits, we get
$\underset{x\to {{4}^{+}}}{\mathop{\lim }}\,f(x)=\sqrt{4-4}$
$\underset{x\to {{4}^{+}}}{\mathop{\lim }}\,f(x)=0.........(ii)$
So the right hand limit exists and is equal to zero.
So from equation (i) and (ii), we get
$\underset{x\to {{4}^{-}}}{\mathop{\lim }}\,f(x)=\underset{x\to {{4}^{+}}}{\mathop{\lim }}\,f(x)$
Therefore, $\underset{x\to 4}{\mathop{\lim }}\,f(x)$exists and is equal to zero.

Note: For finding the left hand limit we applied limits rules, instead of that we can directly apply the limits to find out the left hand limit value.