Question

1. find the limit.

\\( \lim _ { x \
ightarrow 0 } \frac { x ^ { 2 } - x + \sin x } { 2 x } \\)

Explanation:

Step1: Substitute \(x = 0\) into the original expression

Substitute \(x = 0\) into \(\frac{x^{2}-x+\sin x}{2x}\). We get \(\frac{0^{2}-0+\sin(0)}{2\times0}=\frac{0}{0}\), which is an indeterminate form. So, we can apply L'Hopital's Rule.

Step2: Apply L'Hopital's Rule

Differentiate the numerator and the denominator. The derivative of \(x^{2}-x+\sin x\) is \(2x - 1+\cos x\) (using the power rule \((x^{n})^\prime=nx^{n - 1}\), \((\sin x)^\prime=\cos x\)), and the derivative of \(2x\) is \(2\). So, the limit becomes \(\lim_{x
ightarrow0}\frac{2x - 1+\cos x}{2}\).

Step3: Substitute \(x = 0\) into the new - expression

Substitute \(x = 0\) into \(\frac{2x - 1+\cos x}{2}\). We have \(\frac{2\times0-1+\cos(0)}{2}\). Since \(\cos(0) = 1\), then \(\frac{0 - 1+1}{2}=\frac{0}{2}=0\).

Answer:

\(0\)