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Consider the following statements:
1- The function $f(x)=\sqrt[3]{x}$ is continuous at all $x$ except at $x=0$.
2- The function $f(x)=[x]$ is continuous at $x=2.99$ where [.]is the bracket function. Which of the above statements is/are correct?
Options:
1- The function $f(x)=\sqrt[3]{x}$ is continuous at all $x$ except at $x=0$.
2- The function $f(x)=[x]$ is continuous at $x=2.99$ where [.]is the bracket function. Which of the above statements is/are correct?
Solution:
1736 Upvotes
Verified Answer
The correct answer is:
2 only
$\begin{aligned} & \text { LHLf }(2.99-0)=\lim _{h \rightarrow 0}(2.99-h) \\ & \lim _{h \rightarrow 0}(2.99-h)=\lim _{h \rightarrow 0} 2=2 \\ & \text { RHL } f(2.99-0)=\lim _{h \rightarrow 0} f(2.99+h) \\ &=\lim _{h \rightarrow 0}(2.99+h)=\lim _{h \rightarrow 0} 2=2 \\ & \text { LHL }=\text { RHL } \end{aligned}$
$\therefore \mathrm{f}(\mathrm{x})$ is continous at $\mathrm{x}=2.99$
$\therefore \mathrm{f}(\mathrm{x})$ is continous at $\mathrm{x}=2.99$
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