Evaluate the following in terms of α=limx→0sin⁡xx: limx→0tan2(x)+2xx+x2

Question

Evaluate the following in terms of α=limx→0sin⁡xx:  limx→0tan2(x)+2xx+x2

Lakisha Archer · Accepted Answer

Given,  limx→0tan2(x)+2xx+x2  Now taking x common from numerator and denominator,   we get,  limx→0tan2(x)+2xx+x2=tan⁡(x)tan⁡xx+21+x  as we know, limx→0tan⁡xx=1   So,  limx→0tan⁡(x)tan⁡xx+21+x=0⋅1+21+0=2

RizerMix · Accepted Answer

You could write tan2⁡x+2xx+x2=tan2+2xx(x+1)=sin⁡xx⋅sin⁡xcos2⁡x+2x+1 and use limit laws to get the limit α⋅0+21

alenahelenash · Accepted Answer

Since limx→0cos⁡x=1 we have limx→0tan⁡xx=limx→0sin⁡xx=α , i.e. tan⁡x=αx+o(x) So tan2⁡x=α2x2+o(x2) and tan2⁡x+2x=2x+o(x)=x(2+o(1)) Then tan2⁡x+2xx(1+x)=2+o(1)1+x and the limit as x→0 is 2

Evaluate the following in terms of α=limx→0sin⁡xx: limx→0tan2(x)+2xx+x2

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Answer & Explanation