Find the value

Question:

Let $f(x)=\left\{\begin{array}{l}1+x^{2}, 0 \leq x \leq 1 \\ 2-x, x>1\end{array}\right.$

Show that $\lim _{x \rightarrow 1} f(x)$ does not exist.

 

Solution:

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

$\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{-}} 1+x^{2}$

$=1+(1)^{2}$

$=1+1$

$=2$

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

$\lim _{x \rightarrow 1^{+}} f(x)=\lim _{x \rightarrow 1^{+}} 2-x$

$=2-(1)$

$=2-1$

$=1$

$\lim _{x \rightarrow 1^{-}} f(x) \neq \lim _{x \rightarrow 1^{+}} f(x)$

Thus, $\lim _{x \rightarrow 1} f(x)$ does not exist.

 

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