Using L'Hopital Rule, evaluate $$ \lim_{x \to 0} {\left( \frac {1} {x^2}-\frac {\cot x} {x} \right)}$$
I find this question weired.
If we just combine the two terms into one single fraction, we get$$\lim_{x \to 0} {\frac {1-x\cot x} {x^2}}=\frac10=\infty$$
If we follow L'Hopital Rule, this is $\infty-\infty$ form. We follow the following process to convert it into $\frac00$form.$$\infty_1 -\infty_2=\frac 1{\frac 1{\infty_1}}-\frac 1{\frac 1{\infty_2}}=\frac {\frac 1{\infty_2}-\frac 1{\infty_1}}{{\frac 1{\infty_1}}{\frac 1{\infty_2}}}$$
So we will get $$\lim_{x \to 0} {\left( \frac {1} {x^2}-\frac {\cot x} {x} \right)}=\lim_{x \to 0} {\frac {x\tan x-x^2}{x^3\tan x}}$$
If you keep differentiating using the rule you will get rid of the form of $\frac00$ in the third step of differentiation, which give you the answer $1 \over 3$. This method is very tedious. Trust me, you don't want to try.
I am wondering is there a smarter way of solving this question?
Thanks.
Best Answer
A single application of L'Hospital is sufficient:
$$\frac {1} {x^2}-\frac {\cot x} {x}=\frac{\sin x-x\cos x}{x^2\sin x}\xrightarrow{\text{L'Hospital}}\frac{x\sin x}{2x\sin x+x^2\cos x}=\frac{\dfrac{\sin x}x}{2\dfrac{\sin x}x+\cos x}\xrightarrow{\text{sinc}}\frac1{2+1}.$$