PROBLEM
6. Evaluate limx→2π(sinx−cosx)tanx [5 pts]
STEP 1
1. We are given the expression (sinx−cosx)tanx.
2. We need to evaluate the limit as x approaches 2π.
3. Direct substitution might lead to an indeterminate form, so we may need to use L'Hôpital's Rule or other limit properties.
STEP 2
1. Analyze the behavior of sinx−cosx and tanx as x approaches 2π.
2. Determine if the expression (sinx−cosx)tanx results in an indeterminate form.
3. Use logarithmic properties to simplify the expression if necessary.
4. Apply L'Hôpital's Rule or other limit techniques to evaluate the limit.
5. Simplify and find the final value of the limit.
STEP 3
Evaluate sinx−cosx as x→2π:
sin(2π)=1,cos(2π)=0 Thus, sinx−cosx→1−0=1.
STEP 4
Evaluate tanx as x→2π:
tanx=cosxsinx As x→2π, cosx→0, so tanx→∞.
STEP 5
Check if (sinx−cosx)tanx results in an indeterminate form:
Since (sinx−cosx)→1 and tanx→∞, the expression 1∞ is an indeterminate form.
STEP 6
Use logarithmic properties to simplify the expression:
Let y=(sinx−cosx)tanx.
Take the natural logarithm:
lny=tanx⋅ln(sinx−cosx)
STEP 7
Evaluate the limit of lny as x→2π:
x→2πlimlny=x→2πlimtanx⋅ln(sinx−cosx) Since tanx→∞ and ln(sinx−cosx)→ln(1)=0, we have the form ∞⋅0.
Rewrite as a fraction:
x→2πlimcotxln(sinx−cosx)
STEP 8
Apply L'Hôpital's Rule:
Differentiate the numerator and the denominator:
Numerator: dxd[ln(sinx−cosx)]=sinx−cosxcosx+sinx
Denominator: dxd[cotx]=−csc2x
Apply L'Hôpital's Rule:
x→2πlimcotxln(sinx−cosx)=x→2πlim(sinx−cosx)(−csc2x)cosx+sinx
STEP 9
Simplify the expression:
As x→2π, sinx→1 and cosx→0, so the expression simplifies to:
x→2πlim(sinx−cosx)(−csc2x)cosx+sinx=x→2πlim−11=0
SOLUTION
Since limx→2πlny=0, we have:
x→2πlimy=e0=1 The value of the limit is:
1
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