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The limit of [tex]\lim_{x \to -3}\sqrt{ x^2-8}[/tex] as x approaches negative 3 is 1
The limit of the function given in question can also be written as:
[tex]\lim_{x \to -3}\sqrt{ x^2-8}[/tex]
To determine the required limit, we will simply substitute the value of x into the function as shown:
[tex]\lim_{x \to -3}\sqrt{ x^2-8}=\sqrt{ (-3)^2-8}\\\lim_{x \to -3}\sqrt{ x^2-8}=\sqrt{9-8}=1\\[/tex]
Hence the limit of [tex]\lim_{x \to -3}\sqrt{ x^2-8}[/tex] as x approaches negative 3 is 1
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