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Question

Evaluate limx1072x(52)x210

Or

Differentiate x2 cos x by first principle.

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Solution

We have,limx1072x(52)x210

=limx1072x(52)x210

=limx1072x7210x210

=limx1072x7210x210×=limx1072x+721072x+7210

=limx10(72x)(7210)(x+10)(x10)(7x)+7210

=limx102x+210(x+10)(x10)(7x)+7210

=limx102(x10(x+10)(x10)(7x)+7210

limx102(x+10)(72x)+7x10

limx102210(27210)=1210(52)2

=1210(52)=1210(52)×(5+2)(5+2)

=(5+2)210×3=(5+2)610

Or

Let f(x)=x2 cos x.then f(x+h)=(x+h)2 cos (x+h)

f(x)=limh0f(x+h)f(x)h

=limh0(x2+2hx+h2)cos(x+h)x2 cos xh

=limh0x2[cos(x+h)cos x]+h(2x+h)(cos(x+h))h

=limh0x2(2)sin(x+h+x2)sin(x+hx2)h+2x cos x

=limh02x2 sin(x+h2)sinh22×h2+2x cos x

=x2 sin x+2x cos x [=limh0sin hh=1]


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