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What is limit of startroot x squared minus 8 endroot as x approaches negative 3? –5 1 startroot 2 endroot startroot negative 14 endroot

Sagot :

abidemiokin

The limit of [tex]\lim_{x \to -3}\sqrt{ x^2-8}[/tex] as x approaches negative 3 is 1

Limit of a function

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

Learn more on limit here: https://brainly.com/question/1444047
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