Question

use lhôpitals rule to find the following limit.
lim_{x
ightarrow0}\frac{x^{2}}{ln(sec x)}
lim_{x
ightarrow0}\frac{x^{2}}{ln(sec x)}=square \text{ (type an exact answer.)}

Explanation:

Step1: Check indeterminate form

When \(x
ightarrow0\), \(\lim_{x
ightarrow0}x^{2}=0\) and \(\lim_{x
ightarrow0}\ln(\sec x)=\ln(\sec0)=\ln(1) = 0\). So, it is in the \(\frac{0}{0}\) form and l'Hopital's Rule can be applied.

Step2: Differentiate numerator and denominator

The derivative of \(y = x^{2}\) is \(y^\prime=2x\) using the power - rule \((x^n)^\prime=nx^{n - 1}\). The derivative of \(y=\ln(\sec x)\) using the chain - rule: Let \(u = \sec x\), then \(y=\ln(u)\). \(\frac{dy}{du}=\frac{1}{u}\) and \(\frac{du}{dx}=\sec x\tan x\), so \(\frac{d}{dx}\ln(\sec x)=\frac{\sec x\tan x}{\sec x}=\tan x\).
So, \(\lim_{x
ightarrow0}\frac{x^{2}}{\ln(\sec x)}=\lim_{x
ightarrow0}\frac{2x}{\tan x}\).

Step3: Check indeterminate form again

When \(x
ightarrow0\), \(\lim_{x
ightarrow0}2x = 0\) and \(\lim_{x
ightarrow0}\tan x=0\). It is still in the \(\frac{0}{0}\) form, so apply l'Hopital's Rule again.

Step4: Differentiate numerator and denominator again

The derivative of \(y = 2x\) is \(y^\prime=2\), and the derivative of \(y=\tan x\) is \(y^\prime=\sec^{2}x\).
So, \(\lim_{x
ightarrow0}\frac{2x}{\tan x}=\lim_{x
ightarrow0}\frac{2}{\sec^{2}x}\).

Step5: Evaluate the limit

When \(x
ightarrow0\), \(\sec x=\frac{1}{\cos x}\), and \(\cos0 = 1\), so \(\sec0 = 1\). Then \(\lim_{x
ightarrow0}\frac{2}{\sec^{2}x}=\frac{2}{\sec^{2}(0)}=\frac{2}{1^{2}} = 2\).

Answer:

\(2\)