Evaluate the Limit

Problem

(lim_x→0)((3*x−sin(3*x))/(3*x−tan(3*x)))

Solution

1. Identify the indeterminate form by substituting x=0 into the expression.

(3*(0)−sin(0))/(3*(0)−tan(0))=0/0

2. Apply L'Hôpital's Rule by differentiating the numerator and the denominator with respect to x

d(3*x−sin(3*x))/d(x)=3−3*cos(3*x)

d(3*x−tan(3*x))/d(x)=3−3*sec2(3*x)

3. Evaluate the new limit as x→0

(lim_x→0)((3−3*cos(3*x))/(3−3*sec2(3*x)))=(3−3*(1))/(3−3*(1))=0/0

4. Apply L'Hôpital's Rule a second time.

d(3−3*cos(3*x))/d(x)=9*sin(3*x)

d(3−3*sec2(3*x))/d(x)=−18*sec2(3*x)*tan(3*x)

5. Simplify the expression before evaluating or applying the rule again.

(9*sin(3*x))/(−18*sec2(3*x)*tan(3*x))=(9*sin(3*x))/(−18⋅1/cos2(3*x)⋅sin(3*x)/cos(3*x))

(9*sin(3*x))/(−(18*sin(3*x))/cos3(3*x))=−1/2*cos3(3*x)

6. Substitute x=0 into the simplified expression.

−1/2*cos3(3*(0))=−1/2*(1)3=−1/2

Final Answer

(lim_x→0)((3*x−sin(3*x))/(3*x−tan(3*x)))=−1/2

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