Prove the following


$f(x)=\left\{\begin{array}{l}3 x+5, \text { if } x \geq 2 \\ x^{2}, \text { if } x<2\end{array}\right.$ at $\mathrm{x}=2$



Checking the continuity of the given function, we have

$\lim _{x \rightarrow 2^{-}} f(x)=3 x+5$

$=\lim _{h \rightarrow 0} 3(2+h)+5=11$

$\lim _{x \rightarrow 2} f(x)=3 x+5=3(2)+5=11$

$=\lim _{h \rightarrow 0}(2)^{2}+h^{2}-4 h=(2)^{2}=4$

Now, since $\lim _{x \rightarrow 2^{-}} f(x)=\lim _{x \rightarrow 2} f(x) \neq \lim _{x \rightarrow 2} f(x)$

Thus, f(x) is discontinuous at x = 2.

Leave a comment


Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now