Need help with these Hospital limits:
Find: $\displaystyle \lim_{x\to 0}\tan(x)\,\ln(x)$.
Rewrite as:
$\displaystyle \lim_{x\to 0}\frac{\ln(x)}{\cot(x)}$,
now use L'hopital's rule:
$\displaystyle \lim_{x\to 0}\frac{\ln(x)}{\cot(x)}=\lim_{x \to 0}\frac{1/x}{(-\csc^2(x))}$,
...............=$\displaystyle \lim_{x \to 0}-\frac{\sin^2(x)}{x} $
...............=$\displaystyle 0$
RonL
Find $\displaystyle \lim_{x \to 0}\frac{1}{x^2}-\frac{\sin(x)}{x}$
The first term in the limit clearly goes to $\displaystyle \infty$, while the second
term is a well known limit equal to $\displaystyle 1$, so the limit as a whole is $\displaystyle \infty$.
Opps didn't see the actual question:
Find $\displaystyle \lim_{x \to 0}\frac{1}{x^2}-\frac{\cos(x)}{\frac{\sin(x)}{x}}$
but the second term still has a finite limit and so the whle thing still goes to $\displaystyle \infty$
(or is a different interpretation of the second term in the limit meant?)
RonL
Hello, totalnewbie!
Here's the first one . . .
$\displaystyle 1)\;\lim_{x\to0}\left(\tan x\cdot\ln x\right)$
The function goes to: $\displaystyle (0)(-\infty)$
Rewrite as: .$\displaystyle \lim_{x\to0}\frac{\ln x}{\cot x}$ . . . which goes to $\displaystyle \frac{-\infty}{\infty}$
Apply L'Hopital: .$\displaystyle \lim_{x\to0}\left(\frac{\frac{1}{x}}{-\csc^2x}\right)\;=\;\lim_{x\to0}\left(-\frac{\sin^2x}{x}\right)$
. . $\displaystyle = \;\lim_{x\to0}\left[(-\sin x)\left(\frac{\sin x}{x}\right)\right] \;= \;-(0)(1) \;=\;\boxed{0}$
Third go at the second question:
Find $\displaystyle \lim_{x \to 0} \frac{1}{x^2}-\frac{\cos(x)/\sin(x)}{x}$
which seems a plausible form for what might have been intended.
(the limit is 1/3 but I will need to do some work to produce what you would
think an acceptable explanation)
The limit is:
$\displaystyle \lim_{x \to 0} \frac{1}{x^2}-\frac{\cot(x)}{x}$
Take the series expasion of $\displaystyle \cot(x)=\frac{1}{x}-\frac{x}{3}+O(x^3)$
So:
$\displaystyle \lim_{x \to 0} \frac{1}{x^2}-\frac{\cot(x)}{x}=\lim_{x \to 0}\frac{1}{x^2}-(1/x)\left[\frac{1}{x}-\frac{x}{3}+O(x^3)\right]$
..............$\displaystyle =\lim_{x \to 0}\frac{1}{3}-O(x^2)=\frac{1}{3}$
RonL
Hello, totalnewbie!
I am assuming that the second problem is: .$\displaystyle \lim_{x\to0}\frac{1}{x^2} - \frac{\frac{\cos x}{\sin x}}{x} $
Then we have: .$\displaystyle \frac{1}{x^2}-\frac{\cos x}{x\sin x} \;=\;\frac{\sin x - x\cos x}{x^2\sin x}$ . . . which goes to $\displaystyle \frac{0}{0}$
Apply L'Hopital: .$\displaystyle \frac{\cos x + x\sin x - \cos x}{x^2\cos x + 2x\sin x} \:=\:\frac{x\sin x}{x^2\cos x + 2x\sin x}$ $\displaystyle = \:\frac{\sin x}{x\cos x + 2\sin x}$
Divide top and bottom by $\displaystyle \sin x:\;\;\frac{1}{\frac{x}{\sin x}\!\cdot\!\cos x + 2}$
Then: .$\displaystyle \lim_{x\to0}\left(\frac{1}{\frac{x}{\sin x}\!\cdot\!\cos x + 2}\right) \;=\;\frac{1}{1\!\cdot1 + 2} \;=\;\boxed{\frac{1}{3}}$