Given below are two statements: **Statement I:** $$\lim_{x \to 0} \left( \frac{\tan^{-1}x + \log_e \left( \frac{\sqrt{1 + x} - \sqrt{1 - x}}{x} \right) - 2x}{x^5} \right) = \frac{2}{5}$$ **Statement II:** $$\lim_{x \to 1} \left( \frac{2}{x^{1 - x}} \right) = \frac{1}{e^2}$$ In the light of the above statements, choose the correct answer from the options given below:

Step 1: Evaluate Statement I using Taylor series expansions

To evaluate the limit in Statement I, we will use Taylor series expansions for each term around $x=0$. We need to expand up to the $x^5$ term because the denominator is $x^5$. The expansions for $\tan^{-1}x$, $\sqrt{1+x}$, and $\log_e(1+y)$ are shown.

Step 2: Simplify the logarithmic term in Statement I

First, we simplify the argument of the logarithm. We substitute the Taylor expansions for $\sqrt{1+x}$ and $\sqrt{1-x}$ and simplify the expression. Then, we substitute this simplified expression into the Taylor expansion for $\log_e(1+y)$ to get the expansion for the logarithmic term.

Step 3: Substitute expansions into Statement I and evaluate the limit

Now, we substitute the Taylor series expansions for $\tan^{-1}x$ and the logarithmic term back into the original limit expression. We also include the $-2x$ term. After combining like terms, we can see that the numerator does not have an $x^5$ term with a non-zero coefficient that would result in $2/5$. The lowest power of $x$ in the numerator after simplification is $x$, which means the limit will be undefined or zero, not $2/5$. Therefore, Statement I is false.

Step 4: Evaluate Statement II using properties of limits

For Statement II, we need to evaluate the limit as $x \to 1$. We can directly substitute $x=1$ into the expression since the function is continuous at $x=1$. Substituting $x=1$ gives $1^{1-1} = 1^0 = 1$. Therefore, the limit is $2/1 = 2$. Since the statement claims the limit is $1/e^2$, Statement II is false.

More from this chapter

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