Prove that

Question:

Let $f(x)=\left\{\begin{array}{l}4 x-5, x \leq 2 \\ x-a, x>2\end{array}\right.$

If $\lim _{x \rightarrow 2} f(x)$ exists then find the value of $a$.

 

Solution:

Left Hand Limit(L.H.L.):

$\lim _{x \rightarrow 2^{-}} f(x)=\lim _{x \rightarrow 2^{-}} 4 x-5$

$=4(2)-5$

$=8-5$

$=3$

Right Hand Limit(R.H.L.):

$\lim _{x \rightarrow 2^{+}} f(x)=\lim _{x \rightarrow 2^{+}} x-a$

$=2-\mathrm{a}$

Since $\lim _{x \rightarrow 2} f(x)$ it exists,

$\lim _{x \rightarrow 2^{-}} f(x)=\lim _{x \rightarrow 2^{+}} f(x)$

$\rightarrow 3=2-a$

$\rightarrow a=2-3$

$\rightarrow a=-1$

 

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