Question

The value of $\lim _{\mathrm{x} \rightarrow 1}\left(\frac{\mathrm{x}^{3}+2 \mathrm{x}^{2}+\mathrm{x}+1}{\mathrm{x}^{2}+2 \mathrm{x}+3}\right)^{\frac{1-\cos (\mathrm{x}-1)}{(\mathrm{x}-1)^{2}}}$ is (A) e (B) \(\mathrm{e}^{1 / 2}\) (C) 1 (D) none of these

Short answer

Question: Determine the limit of the given expression: \(\lim_{x\to1}\left(\frac{x^{3}+2x^{2}+x+1}{x^{2}+2x+3}\right)^{\frac{1-\cos(x-1)}{(x-1)^2}}\) Answer: (B) \(\left(\frac{5}{6}\right)^{\frac{1}{2}}\)

Step-by-step solution

Find the limit of the function inside the parentheses

We have \(\lim _{x \rightarrow 1}\left(\frac{x^{3}+2x^{2}+x+1}{x^{2}+2x+3}\right)\). We can substitute x = 1 into the expression and simplify: \(\frac{1^{3}+2(1)^{2}+1+1}{1^{2}+2(1)+3}=\frac{1+2+1+1}{1+2+3}=\frac{5}{6}\) So, the limit of the function inside the parentheses as x approaches 1 is \(\frac{5}{6}\).

Find the limit of the exponent

The given exponent is \(\frac{1 - \cos(x-1)}{(x-1)^2}\). Let \(u = x - 1\). Then as \(x \to 1\), \(u \to 0\). Therefore, we have: \(\lim_{x\to1}\frac{1 - \cos(x-1)}{(x-1)^2} = \lim_{u\to0}\frac{1 - \cos(u)}{u^2}\) The standard limit \(\lim_{u\to0}\frac{1 - \cos(u)}{u^2} = \frac{1}{2}\) is a known result. So, the limit of the exponent as x approaches 1 is \(\frac{1}{2}\).

Use the limit laws to find the limit of the given expression

Using the limit laws, we can find the limit of the given expression: \(\lim_{x\to1}\left(\frac{x^{3}+2x^{2}+x+1}{x^{2}+2x+3}\right)^{\frac{1-\cos(x-1)}{(x-1)^2}}\) \(=\left(\lim_{x\to1}\frac{x^{3}+2x^{2}+x+1}{x^{2}+2x+3}\right)^{\lim_{x\to1}\frac{1-\cos(x-1)}{(x-1)^2}}\) \(= \left(\frac{5}{6}\right)^{\frac{1}{2}}\) So, the limit of the given expression is \(\left(\frac{5}{6}\right)^{\frac{1}{2}}\), which matches option (B) \(\mathrm{e}^{1 / 2}\) as the final answer.