Question Number 162726 by LEKOUMA last updated on 31/Dec/21
![1) Calculate lim_(x→0) ((tgx^m )/((sin x)^n )), (m, n∈ N) 2) f′(a) existe, calculate lim_(x→+∞) x[f(a+(a/x))−f(a−(β/x))], (α, β ∈ R)](Q162726.png)
$$\left.\mathrm{1}\right)\:{Calculate} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{tgx}^{{m}} }{\left(\mathrm{sin}\:{x}\right)^{{n}} },\:\:\left({m},\:{n}\in\: {N}\right) \\ $$$$\left.\mathrm{2}\right)\:{f}'\left({a}\right)\:{e}\mathrm{xiste},\:{calculate} \\ $$$$\underset{{x}\rightarrow+\infty} {\mathrm{lim}}{x}\left[{f}\left({a}+\frac{{a}}{{x}}\right)−{f}\left({a}−\frac{\beta}{{x}}\right)\right],\: \\ $$$$\left(\alpha,\:\beta\:\in\: {R}\right) \\ $$
Answered by mr W last updated on 01/Jan/22
![(1) lim_(x→0) ((tgx^m )/((sin x)^n )) =lim_(x→0) ((tgx^m )/x^m )×((x/(sin x)))^n ×x^(m−n) =lim_(x→0) x^(m−n) = { ((1 if m=n)),((0 if m>n)),((∞ if m<n)) :} (2) lim_(x→+∞) x[f(a+(α/x))−f(a)−f(a−(β/x))+f(a)] =lim_(x→+∞) [((f(a+(α/x))−f(a))/(1/x))−((f(a−(β/x))−f(a))/(1/x))] =lim_(x→+∞) [α×((f(a+(α/x))−f(a))/(α/x))+β×((f(a−(β/x))−f(a))/(−(β/x)))] =lim_(h,k→0) [α×((f(a+h)−f(a))/h)+β×((f(a+k)−f(a))/k)] =αf′(a)+βf′(a) =(α+β)f′(a)](Q162745.png)
$$\left(\mathrm{1}\right) \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{tgx}^{{m}} }{\left(\mathrm{sin}\:{x}\right)^{{n}} } \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{tgx}^{{m}} }{{x}^{{m}} }×\left(\frac{{x}}{\mathrm{sin}\:{x}}\right)^{{n}} ×{x}^{{m}−{n}} \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}{x}^{{m}−{n}} =\begin{cases}{\mathrm{1}\:{if}\:{m}={n}}\\{\mathrm{0}\:{if}\:{m}>{n}}\\{\infty\:{if}\:{m}<{n}}\end{cases} \\ $$$$ \\ $$$$\left(\mathrm{2}\right) \\ $$$$\underset{{x}\rightarrow+\infty} {\mathrm{lim}}{x}\left[{f}\left({a}+\frac{\alpha}{{x}}\right)−{f}\left({a}\right)−{f}\left({a}−\frac{\beta}{{x}}\right)+{f}\left({a}\right)\right]\: \\ $$$$=\underset{{x}\rightarrow+\infty} {\mathrm{lim}}\left[\frac{{f}\left({a}+\frac{\alpha}{{x}}\right)−{f}\left({a}\right)}{\frac{\mathrm{1}}{{x}}}−\frac{{f}\left({a}−\frac{\beta}{{x}}\right)−{f}\left({a}\right)}{\frac{\mathrm{1}}{{x}}}\right]\: \\ $$$$=\underset{{x}\rightarrow+\infty} {\mathrm{lim}}\left[\alpha×\frac{{f}\left({a}+\frac{\alpha}{{x}}\right)−{f}\left({a}\right)}{\frac{\alpha}{{x}}}+\beta×\frac{{f}\left({a}−\frac{\beta}{{x}}\right)−{f}\left({a}\right)}{−\frac{\beta}{{x}}}\right]\: \\ $$$$=\underset{{h},{k}\rightarrow\mathrm{0}} {\mathrm{lim}}\left[\alpha×\frac{{f}\left({a}+{h}\right)−{f}\left({a}\right)}{{h}}+\beta×\frac{{f}\left({a}+{k}\right)−{f}\left({a}\right)}{{k}}\right]\: \\ $$$$=\alpha{f}'\left({a}\right)+\beta{f}'\left({a}\right) \\ $$$$=\left(\alpha+\beta\right){f}'\left({a}\right) \\ $$
Commented by Ar Brandon last updated on 01/Jan/22
👏👏👏First comment of the year (GMT+1) Happy New year, Sir !
Commented by mr W last updated on 01/Jan/22
$${thanks}! \\ $$$${the}\:{same}\:{to}\:{you}\:{and}\:{all}\:{others}! \\ $$
Commented by Ar Brandon last updated on 01/Jan/22
$$\mathrm{Thanks}\:\mathrm{Sir}. \\ $$