How do you solve for L in T = 2 π √ L g ?
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To solve for ( L ) in the equation ( T = 2\pi \sqrt{L/g} ), where ( T ) represents the period of oscillation, and ( g ) is the acceleration due to gravity, follow these steps:

Square both sides of the equation to isolate the square root term: ( T^2 = 4\pi^2 \frac{L}{g} ).

Multiply both sides of the equation by ( g ) to move ( L ) out of the denominator: ( T^2 g = 4\pi^2 L ).

Divide both sides of the equation by ( 4\pi^2 ) to isolate ( L ): ( L = \frac{T^2 g}{4\pi^2} ).

This expression gives the value of ( L ) in terms of ( T ) and ( g ).
Therefore, ( L = \frac{T^2 g}{4\pi^2} ).
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When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
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