Question

The right hand and left hand limit of the function
f(x) = \(\left\{ \begin{aligned}     & \frac{e^{1/x}-1}{e^{1/x}+1}\ \ ,\ \ \text{if x}\ne0   \\     & \ \ \ \ \ \ 0 \ \ \ \ \ \ \ \ ,\ \ \text{if x = 0}    \end{aligned} \right.\) are respectively

• A. 1 and 1
• B. 1 and -1
• C. -1 and -1
• D. -1 and 1

Answer 1

The Correct Option is B

We need to find the right-hand limit and left-hand limit of the function:

\(f(x) = \begin{cases} \frac{e^{1/x} - 1}{e^{1/x} + 1}, & \text{if } x \neq 0 \\ 0, & \text{if } x = 0 \end{cases}\)

Right-hand limit (x approaches 0 from the right):

\(\lim_{x \to 0^+} \frac{e^{1/x} - 1}{e^{1/x} + 1}\)

As \(x \to 0^+\), \(\frac{1}{x} \to \infty\), so \(e^{1/x} \to \infty\).

We can divide both the numerator and denominator by \(e^{1/x}\):

\(\lim_{x \to 0^+} \frac{1 - e^{-1/x}}{1 + e^{-1/x}}\)

As \(x \to 0^+\), \(\frac{1}{x} \to \infty\), so \(e^{-1/x} \to 0\).

\(\lim_{x \to 0^+} \frac{1 - 0}{1 + 0} = \frac{1}{1} = 1\)

Left-hand limit (x approaches 0 from the left):

\(\lim_{x \to 0^-} \frac{e^{1/x} - 1}{e^{1/x} + 1}\)

As \(x \to 0^-\), \(\frac{1}{x} \to -\infty\), so \(e^{1/x} \to 0\).

\(\lim_{x \to 0^-} \frac{0 - 1}{0 + 1} = \frac{-1}{1} = -1\)

So, the right-hand limit is 1 and the left-hand limit is -1.

Therefore, the correct option is (B) 1 and -1.

Answer 2

We are given the function:

$$ f(x) = \begin{cases} \frac{e^{1/x} - 1}{e^{1/x} + 1}, & \text{if } x \neq 0 \\ 0, & \text{if } x = 0 \end{cases} $$

We need to find the right-hand and left-hand limits of the function as $ x \to 0 $.

When $ x \to 0^+ $ (from the right), $ \frac{1}{x} \to +\infty $. Thus, we analyze the function:

$$ f(x) = \frac{e^{1/x} - 1}{e^{1/x} + 1} $$

As $ x \to 0^+ $, $ e^{1/x} \to \infty $. Hence, the expression becomes:

$$ f(x) \approx \frac{\infty - 1}{\infty + 1} \to \frac{\infty}{\infty} \to 1 $$

Thus, the right-hand limit is $ \lim_{x \to 0^+} f(x) = 1 $.

When $ x \to 0^- $ (from the left), $ \frac{1}{x} \to -\infty $. Now we analyze the function:

$$ f(x) = \frac{e^{1/x} - 1}{e^{1/x} + 1} $$

As $ x \to 0^- $, $ e^{1/x} \to 0 $. Thus, the expression becomes:

$$ f(x) \approx \frac{0 - 1}{0 + 1} = \frac{-1}{1} = -1 $$

Thus, the left-hand limit is $ \lim_{x \to 0^-} f(x) = -1 $.

Answer: (B) $ 1 $ and $ -1 $.