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1] f) Evaluate $(\lim)_{x->0}\frac{e^{2x} - (1+x)^2}{xlog(1+x)}$

written 7.5 years ago by teamques10 ★ 70k

•   modified 7.5 years ago

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written 7.5 years ago by teamques10 ★ 70k

$lim_{x→0}⁡1.⁡\frac{(e^{2x} - (1+x)^2)}{\frac{(xlog(1+x)}{x}.x}$ = $lim_{x→0}⁡1.⁡\frac{(e^{2x} - (1+x)^2)}{x}.x$ = $lim_{x→0}⁡1.\frac{(e^{2x} - (1+x)^2)}{x^2}$ Applying L-Hospital rule $lim_{x→0}⁡1.\frac{(2e^{2x} - 2(1+x)^1)}{2x} = lim_{x→0}⁡1.\frac{(4e^{2x} – 2)}{2} =\frac{(4e^0-2)}{2}= 1.$

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