Question

1. $limlimits_{x \to 0} \frac{\tan^{-1}(2x)}{3x}$

Explanation:

Step1: Apply L'Hopital's Rule

When \(x
ightarrow0\), \(\tan^{- 1}(2x)
ightarrow0\) and \(3x
ightarrow0\). So, we can use L'Hopital's Rule. The derivative of \(\tan^{-1}(u)\) with respect to \(x\) is \(\frac{u'}{1 + u^{2}}\) (by the chain - rule). Here \(u = 2x\), so \(u'=2\). The derivative of \(\tan^{-1}(2x)\) is \(\frac{2}{1+(2x)^{2}}\), and the derivative of \(3x\) is \(3\).

$$ \lim_{x ightarrow0}\frac{\tan^{-1}(2x)}{3x}=\lim_{x ightarrow0}\frac{\frac{2}{1 + 4x^{2}}}{3} $$

Step2: Evaluate the limit

As \(x
ightarrow0\), substitute \(x = 0\) into \(\frac{\frac{2}{1 + 4x^{2}}}{3}\).

$$ \frac{\frac{2}{1+4\times0^{2}}}{3}=\frac{2}{3} $$

Answer:

\(\frac{2}{3}\)