Question

in which limits below can we use lhospitals rule?
\\(\square \lim\limits_{x\to \pi/4} \dfrac{\sin(4x)}{4x - \pi}\\)
\\(\square \lim\limits_{x\to 0} \dfrac{1 - e^x}{\sin(3x)}\\)
\\(\square \lim\limits_{x\to 0} \dfrac{3x}{\cot x}\\)
\\(\square \lim\limits_{x\to \infty} \dfrac{\ln x}{\sqrt{x}}\\)

Explanation:

Step1: Check the form of each limit

• For \(\lim_{x

ightarrow\pi/4}\frac{\sin(4x)}{4x - \pi}\):
Substitute \(x = \frac{\pi}{4}\), \(\sin(4\times\frac{\pi}{4})=\sin(\pi) = 0\) and \(4\times\frac{\pi}{4}-\pi=0\). So it is in \(\frac{0}{0}\) form.

• For \(\lim_{x

ightarrow0}\frac{1 - e^{x}}{\sin(3x)}\):
Substitute \(x = 0\), \(1 - e^{0}=0\) and \(\sin(0) = 0\). So it is in \(\frac{0}{0}\) form.

• For \(\lim_{x

ightarrow0}\frac{3x}{\cot x}\):
Substitute \(x = 0\), \(3\times0 = 0\) and \(\cot(0)=\infty\). But rewrite \(\frac{3x}{\cot x}=3x\tan x\). Substitute \(x = 0\), \(3\times0\times0 = 0\). Also, using L'Hospital's Rule, \(\lim_{x
ightarrow0}\frac{3x}{\cot x}=\lim_{x
ightarrow0}\frac{3}{\frac{-1}{\sin^{2}x}}=\lim_{x
ightarrow0}- 3\sin^{2}x = 0\). It is in \(\frac{0}{0}\) form (after rewriting \(\cot x=\frac{\cos x}{\sin x}\), \(\frac{3x}{\cot x}=\frac{3x\sin x}{\cos x}\), and when \(x = 0\), \(3\times0\times0/1 = 0\)).

• For \(\lim_{x

ightarrow\infty}\frac{\ln x}{\sqrt{x}}\):
As \(x
ightarrow\infty\), \(\ln x
ightarrow\infty\) and \(\sqrt{x}
ightarrow\infty\). So it is in \(\frac{\infty}{\infty}\) form.

Answer:

All of the given limits \(\lim_{x
ightarrow\pi/4}\frac{\sin(4x)}{4x - \pi}\), \(\lim_{x
ightarrow0}\frac{1 - e^{x}}{\sin(3x)}\), \(\lim_{x
ightarrow0}\frac{3x}{\cot x}\), \(\lim_{x
ightarrow\infty}\frac{\ln x}{\sqrt{x}}\) can use L'Hospital's Rule.