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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