lim = infinity 2+sinx / x^2 + 3
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Given, x→0limx2sinxx−sinx=x→0limx3x−sinx.sinxx
=x→0limx3x−sinx.1 (∵x→0limxsinx=1)
Applying L' Hospital's Rule, we have
=x→0limx3x−sinx ...00 form
=x→0lim3x21−cosx ...00 form
=x→0lim6xsinx ...00 form
=61x→0limxsinx=61.
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