Question

use lhôpitals rule to find the limit.

$$\lim_{x \to 0} \frac{8x^{2}}{\cos(x) - 1}$$

$$\lim_{x \to 0} \frac{8x^{2}}{\cos(x) - 1} = \square$$ (type an exact answer.)

Explanation:

Step1: Check the form

When \(x = 0\), \(\frac{8x^{2}}{\cos(x)-1}=\frac{0}{0}\) (indeterminate form), so apply L'Hôpital's Rule.
Differentiate numerator and denominator.
The derivative of \(y = 8x^{2}\) is \(y^\prime=16x\) (using the power rule \((x^{n})^\prime=nx^{n - 1}\)).
The derivative of \(y=\cos(x)-1\) is \(y^\prime=-\sin(x)\) (using \((\cos x)^\prime=-\sin x\) and \((c)^\prime = 0\) for constant \(c\)).
So, \(\lim_{x
ightarrow0}\frac{8x^{2}}{\cos(x)-1}=\lim_{x
ightarrow0}\frac{16x}{-\sin(x)}\).

Step2: Re - check the form and apply L'Hôpital's Rule again

When \(x = 0\), \(\frac{16x}{-\sin(x)}=\frac{0}{0}\) (indeterminate form).
Differentiate numerator and denominator again.
The derivative of \(y = 16x\) is \(y^\prime=16\).
The derivative of \(y=-\sin(x)\) is \(y^\prime=-\cos(x)\).
So, \(\lim_{x
ightarrow0}\frac{16x}{-\sin(x)}=\lim_{x
ightarrow0}\frac{16}{-\cos(x)}\).

Step3: Evaluate the limit

Substitute \(x = 0\) into \(\frac{16}{-\cos(x)}\).
When \(x = 0\), \(\cos(0)=1\). So \(\frac{16}{-\cos(0)}=- 16\).

Answer:

\(-16\)