1.

`lim_(xto(pi//4)) (tan^(3)x-tanx)/(cos(x+pi/4))`

Answer» Given, `underset(xto(pi//4))"lim"(tan^(3)x-tanx)/(cos(x+pi/4))`
`=underset(xto(pi)/4)"lim"(tanx(tan^(2)x-1))/(cos(x+pi/4))= underset(xto(pi//4))"lim"tanx.underset(xto(pi//4))"lim"((1-tan^(2)x)/(cos(x+(gpi)/4)))`
`=underset(xto(pi//4))(-1xx"lim")((1+tanx)(1-tanx))/(cos(x+pi/4))` `[therefore a^(2)-b^(2)=(a+b)(a-b)]`
`-underset(xto(pi//4))"lim"(1+tanx) underset(xto(pi//4))"lim"[(cosx-sinx)/(cosx.cos(x+pi/4)]]`
`=-(1+1) xx underset(xto(pi//4))"lim"(sqrt(2)[1/sqrt(2).cosx-1/sqrt(2).sinx])/(cosx.cos(x+pi/4))=-2sqrt(2)underset(xto(pi//4))"lim"[(cospi/4.cosx-sinpi/4.sinx)/(cosx.cos(x+pi/4))]` `[therefore cosA.cosB-sinAsinB=cos(A+B)]`
`=-2sqrt(2)underset(xto(pi//4))"lim"(cos(x+pi/4))/(cos.cos(x+pi/4))=-2sqrt(2)xx1/(1/sqrt(2))=-2sqrt(2)xx sqrt(2)=-4`


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