Question

evaluate the following limit. use lhopitals rule when it is convenient and applicable.
lim_{x
ightarrow0}\frac{1 - cos5x}{4x^{2}}
lim_{x
ightarrow0}\frac{1 - cos5x}{4x^{2}}=square \text{ (type an exact answer.)}

Explanation:

Step1: Check form of limit

When $x\to0$, we have $\frac{1 - \cos(5x)}{4x^{2}}=\frac{1-\cos(0)}{0}=\frac{1 - 1}{0}=\frac{0}{0}$, so L'Hopital's Rule is applicable.

Step2: Differentiate numerator and denominator

The derivative of $1-\cos(5x)$ with respect to $x$ is $5\sin(5x)$ (by chain - rule, derivative of $\cos(u)$ is $-\sin(u)$ and $u = 5x$ so $\frac{d}{dx}(1-\cos(5x))=0-(-\sin(5x))\times5 = 5\sin(5x)$), and the derivative of $4x^{2}$ with respect to $x$ is $8x$. So the limit becomes $\lim_{x\to0}\frac{5\sin(5x)}{8x}$.

Step3: Check form again and apply L'Hopital's Rule if needed

When $x\to0$, $\frac{5\sin(5x)}{8x}=\frac{0}{0}$. Differentiating the numerator and denominator again. The derivative of $5\sin(5x)$ with respect to $x$ is $25\cos(5x)$ (by chain - rule, derivative of $\sin(u)$ is $\cos(u)$ and $u = 5x$ so $\frac{d}{dx}(5\sin(5x))=5\times\cos(5x)\times5=25\cos(5x)$), and the derivative of $8x$ with respect to $x$ is $8$. So the new limit is $\lim_{x\to0}\frac{25\cos(5x)}{8}$.

Step4: Evaluate the limit

Substitute $x = 0$ into $\frac{25\cos(5x)}{8}$. We get $\frac{25\cos(0)}{8}=\frac{25\times1}{8}=\frac{25}{8}$.

Answer:

$\frac{25}{8}$