I'm not sure if my solution is correct. The limit is:
$\lim_{x\to0}\cot(x)-\frac{1}{x}$. Here is how I tried to solve it:
1. $\lim_{x\to0}\cot(x)-\frac{1}{x}$ = $\lim_{x\to0}\cot(x) - x^{-1}$
2. Since $\cot(0)$ is not valid, apply the de l'Hôpital rule $(\cot (x) )' = -\frac{1}{\sin^2 x}=\sin^{-2} x $ and $(x^{-1})'$ = $x^{-2}$
3. $\lim_{x\to0}\sin^{-2} (x)-x^{-2} = 0 - 0 = 0$
However I'm not sure that my logic is correct
Trouble calcualting simple limit
1
$\begingroup$
calculus
limits
-
0no, it's not correct – 2017-01-18
-
1(1) ....$(x^{-1})'=-x^{-2},$ not $+x^{-2}$.... (2) Neither $(\sin x)^{-2}$ nor $x^{-2}$ converges to $0$ as $x\to 0$....(3). $\lim_{x\to 0} ((\sin x)^{-2}-x^{-2})=-1/3$...(4) l'Hopital's rule is for a RATIO of two functions, not their difference..... – 2017-01-18
3 Answers
0
\begin{align} \lim_{x \rightarrow 0} cot(x)-x^{-1} &= \lim_{x \rightarrow 0} \frac{x\cos (x)-\sin(x)}{x\sin(x)} \\ &= \lim_{x \rightarrow 0} \frac{\cos(x)-x\sin(x)-\cos(x)}{\sin(x)+x\cos(x)}, \text{ L'Hôpital's} \\ &= \lim_{x \rightarrow 0} \frac{-x\sin(x)}{\sin(x)+x\cos(x)} \\ &= \lim_{x \rightarrow 0} \frac{-\sin(x)-x \cos(x)}{\cos(x)-x\sin(x)+\cos(x)}\text{, L'Hôpital's} \\&=0 \end{align}
0
For the first one, $\tan x \sim x$ for small $x$ by Taylor's yields $\cot x\sim 1/x$ which should make the limit pretty easy.
To apply L'Hôpital's I always convert it to the indeterminate form $\frac{\infty}{\infty}$ to avoid mistakes. Although you are justified in applying L'Hôpital's in this case as $\infty-\infty$ is an indeterminate form.
-
0The implementation of L'Hopital's isn't correct however – 2017-01-18
-
0@ZacharySelk possible, I did not check this. My apologies – 2017-01-18
0
Hint:
$$\cot(x)-\frac1x=\frac{\cos(x)}{\sin(x)}-\frac1x=\frac{x\cos(x)-\sin(x)}{x\sin(x)}$$
Now try L'Hospital's rule, as it is now of the form $0/0$.
-
0I got it. However, I don't understand is it necessary to have $0/0$ or $\infty /\infty $ to apply L'Hospital's rule? – 2017-01-18
-
0yes It is one of the hypotheses of Theorem – 2017-01-18
-
0@nameLess Yes, you need that to do L'Hospital's rule. – 2017-01-18