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Question Number 122192 by physicstutes last updated on 14/Nov/20

$$\mathrm{Prove}\mathrm{using}\mathrm{the}\mathrm{squeeze}\mathrm{law}\mathrm{of}\mathrm{functions}\mathrm{that}$$$$\underset{x\to a}{\lim }\sqrt{x}=\sqrt{a}.$$$$\mathrm{Recall}\mathrm{that}\mathrm{the}\mathrm{squeeze}\mathrm{theorem}\mathrm{states}\mathrm{that}\mathrm{if}$$$$f\left(x\right)⩽\mathrm{g}\left(x\right)⩽h\left(x\right)$$$$\mathrm{and}\underset{x-a}{\lim }f\left(x\right)=\underset{x\to a}{\lim }h\left(x\right)=L$$$$\mathrm{then},\underset{x\to a}{\lim }\mathrm{g}\left(x\right)=L.$$

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