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ques1 lim [ 4 1/n -1/3 1/n -1] n_# ques2 lim e tanx -e x /tanx-x n_0 ques1 lim [ 41/n-1/31/n-1] n_#ques2 lim etanx-ex/tanx-x n_0
1.0/0 form L hospital rulelim n_# 41/n(-1/n2)ln4/ 31/n(-1/n2)ln3=ln4/ln32. write series expansion of e^xlim n_0 (1+tanx+...)-(1+x+...)/(tanx-x)lim n_0 (tanx-x)+....../(tanx-x)rest of the term will be zero because a^n-b^n has factor a-b alwaysso value of limit is 1Sher Mohammadfaculty askiitians
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