# How to you find the general solution of #(dr)/(ds)=0.05s#?

Now we integrate both sides to undo the differentials:

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To find the general solution of ( \frac{{dr}}{{ds}} = 0.05s ), integrate both sides with respect to ( s ).

[ \int \frac{{dr}}{{ds}} , ds = \int 0.05s , ds ]

[ r = \int 0.05s , ds ]

[ r = 0.05 \times \frac{{s^2}}{2} + C ]

where ( C ) is the constant of integration. Thus, the general solution is ( r = 0.025s^2 + C ).

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When evaluating a one-sided 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.

When evaluating a one-sided 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.

When evaluating a one-sided 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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