Find the limit \(\displaystyle\lim_{{{x}\to{2}}}{\frac{{{{\sin}^{{2}}{\left({x}^{{2}}-{4}\right)}}{{\sec}^{{2}}{\left({3}{x}-{6}\right)}}}}{{{\left({x}^{{3}}-{8}\right)}{\tan{{\left({2}{x}-{4}\right)}}}}}}\)

Jazmin Strong

Jazmin Strong

Answered question

2022-04-15

Find the limit
limx2sin2(x24)sec2(3x6)(x38)tan(2x4)

Answer & Explanation

tabuevniru8op

tabuevniru8op

Beginner2022-04-16Added 14 answers

A little prep work gives us
sin2(x24)sec2(3x6)(x38)tan(2x4)=sin2(x24)(x2)sin(2x4)cos(2x4)(x2+2x+4)cos2(3x6)
Now the second term causes no trouble as x2. It gives
limx2cos(2x4)(x2+2x+4)cos2(3x6)=cos012cos20=112
If we rewrite the first term as
sin2(x24)(x2)sin(2x4)=2(x+2)2(x2)sin2(x24)2(x+2)2(x2)2sin(2x4)
=(x+2)22(sin(x24)(x24))22x4sin(2x4)
we can now use the limit
limu0sinuu=1
with u=x24 for one piece and u=2x4 for another to get
limx2sin2(x24)(x2)sin(2x4)=4221211=8
Combining all this, we have
limx2sin2(x24)sec2(3x6)(x38)tan(2x4)=8112=23

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