Question Number 112851 by bemath last updated on 10/Sep/20
$$\mathrm{Without}\:\mathrm{L}'\mathrm{Hopital} \\ $$$$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left(\frac{\mathrm{a}}{\mathrm{x}}\:−\:\mathrm{cot}\:\frac{\mathrm{x}}{\mathrm{a}}\right)\:? \\ $$
Answered by bobhans last updated on 10/Sep/20
![set (x/a) = t →(a/x) = (1/t) L= lim_(t→0) ((1/t) − cot t) = lim_(t→0) ((1/t)−(1/(tan t))) lim_(t→0) (((tan t−t )/(t.tan t))) . [ let t = 2w ] L = lim_(w→0) ((tan 2w−2w)/(2w.tan 2w)) = lim_(w→0) ((((2tan w)/(1−tan^2 w)) −2w)/(2w.(((2tan w)/(1−tan^2 w))))) L = lim_(w→0) ((2tan w−2w+2wtan^2 w)/(4w.tan w)) L = lim_(w→0) ((tan w−w)/(2w.tan w)) + lim_(w→0) ((2wtan^2 w)/(4wtan w)) L = (1/2)L + 0 ⇒ L=0](https://www.tinkutara.com/question/Q112853.png)
$$\:\mathrm{set}\:\frac{\mathrm{x}}{\mathrm{a}}\:=\:\mathrm{t}\:\rightarrow\frac{\mathrm{a}}{\mathrm{x}}\:=\:\frac{\mathrm{1}}{\mathrm{t}} \\ $$$$\:\mathrm{L}=\:\underset{\mathrm{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left(\frac{\mathrm{1}}{\mathrm{t}}\:−\:\mathrm{cot}\:\mathrm{t}\right)\:=\:\underset{\mathrm{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left(\frac{\mathrm{1}}{\mathrm{t}}−\frac{\mathrm{1}}{\mathrm{tan}\:\:\mathrm{t}}\right) \\ $$$$\underset{\mathrm{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{tan}\:\:\mathrm{t}−\mathrm{t}\:}{\mathrm{t}.\mathrm{tan}\:\mathrm{t}}\right)\:.\:\left[\:\mathrm{let}\:\mathrm{t}\:=\:\mathrm{2w}\:\right] \\ $$$$\mathrm{L}\:=\:\underset{\mathrm{w}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{tan}\:\mathrm{2w}−\mathrm{2w}}{\mathrm{2w}.\mathrm{tan}\:\mathrm{2w}}\:=\:\underset{\mathrm{w}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\frac{\mathrm{2tan}\:\mathrm{w}}{\mathrm{1}−\mathrm{tan}^{\mathrm{2}} \:\mathrm{w}}\:−\mathrm{2w}}{\mathrm{2w}.\left(\frac{\mathrm{2tan}\:\mathrm{w}}{\mathrm{1}−\mathrm{tan}\:^{\mathrm{2}} \mathrm{w}}\right)} \\ $$$$\mathrm{L}\:=\:\underset{\mathrm{w}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{2tan}\:\mathrm{w}−\mathrm{2w}+\mathrm{2wtan}\:^{\mathrm{2}} \mathrm{w}}{\mathrm{4w}.\mathrm{tan}\:\mathrm{w}} \\ $$$$\mathrm{L}\:=\:\underset{\mathrm{w}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{tan}\:\mathrm{w}−\mathrm{w}}{\mathrm{2w}.\mathrm{tan}\:\mathrm{w}}\:+\:\underset{\mathrm{w}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{2wtan}\:^{\mathrm{2}} \mathrm{w}}{\mathrm{4wtan}\:\mathrm{w}} \\ $$$$\mathrm{L}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{L}\:+\:\mathrm{0}\:\Rightarrow\:\mathrm{L}=\mathrm{0} \\ $$
Answered by Dwaipayan Shikari last updated on 10/Sep/20
$$\frac{{a}}{{x}}−{cot}\frac{{x}}{{a}}=\frac{\frac{{a}}{{x}}{tan}\frac{{x}}{{a}}−\mathrm{1}}{{tan}\frac{{x}}{{a}}}=\frac{\frac{{a}}{{x}}\:\frac{{sin}\frac{{x}}{{a}}}{{cos}\frac{{x}}{{a}}}−\mathrm{1}}{\frac{{x}}{{a}}}=\frac{{a}}{{x}}.\frac{\mathrm{1}−{cos}\frac{{x}}{{a}}}{{cos}\frac{{x}}{{a}}} \\ $$$$=\frac{{a}}{{x}}.\frac{\mathrm{2}{sin}^{\mathrm{2}} \frac{{x}}{{a}}}{\mathrm{1}}=\mathrm{2}.\frac{{a}}{{x}}.\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }=\frac{\mathrm{2}{x}}{{a}}=\mathrm{0} \\ $$