Question

use lhopitals rule to find the limit.
lim 8x²
x→0 cos(x) - 1
lim 8x²
x→0 cos(x) - 1 = (type an exact answer.)

Explanation:

Step1: Check the form

When \(x = 0\), \(\frac{8x^{2}}{\cos(x)-1}=\frac{0}{0}\). So, apply L'Hôpital's Rule.

Step2: Differentiate numerator and denominator

Differentiate \(y = 8x^{2}\), \(y^\prime=16x\). Differentiate \(y=\cos(x)-1\), \(y^\prime=-\sin(x)\). Now, \(\lim_{x
ightarrow0}\frac{8x^{2}}{\cos(x)-1}=\lim_{x
ightarrow0}\frac{16x}{-\sin(x)}\).

Step3: Check the new form

When \(x = 0\), \(\frac{16x}{-\sin(x)}=\frac{0}{0}\). Apply L'Hôpital's Rule again.

Step4: Differentiate numerator and denominator again

Differentiate \(y = 16x\), \(y^\prime=16\). Differentiate \(y=-\sin(x)\), \(y^\prime=-\cos(x)\). Now, \(\lim_{x
ightarrow0}\frac{16x}{-\sin(x)}=\lim_{x
ightarrow0}\frac{16}{-\cos(x)}\).

Step5: Substitute \(x = 0\)

Substitute \(x = 0\) into \(\frac{16}{-\cos(x)}\), we get \(\frac{16}{-\cos(0)}=- 16\).

Answer:

\(-16\)