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Question:

Let $f(x)=\left\{\begin{array}{l}\cos x, x \geq 0 \\ x+k, x<0\end{array}\right.$

Find the value of $k$ for which $\lim _{x \rightarrow 0} f(x)$ exist.

 

Solution:

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

$\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{-}} x+k$

$=0+k$

$=k$

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

$\lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0^{+}} \cos x$

$=\cos (0)$

$=1$

It is given that $\lim _{x \rightarrow 0} f(x)$ exists. Therefore,

$\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{+}} f(x)$

$\rightarrow \mathrm{k}=1$

 

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