Question

Let $f(x) = \frac{e^{2x}-1-2x}{x^2}$. Find $\lim_{x\to 0} f(x)$.

• A. 1
• B. 2
• C. 0
• D. 4

Answer 1

Answer: (B)

2

Answer 2

To find $\lim_{x\to 0} \frac{e^{2x}-1-2x}{x^2}$, we first check the form of the limit. Substituting $x=0$ gives $\frac{e^0-1-0}{0^2} = \frac{1-1}{0} = \frac{0}{0}$, which is an indeterminate form.

We can use L'Hopital's Rule.

First application:

• Differentiate numerator: $\frac{d}{dx}(e^{2x}-1-2x) = 2e^{2x}-2$.
• Differentiate denominator: $\frac{d}{dx}(x^2) = 2x$.

The limit becomes $\lim_{x\to 0} \frac{2e^{2x}-2}{2x}$. This is still $\frac{0}{0}$ at $x=0$.

Second application of L'Hopital's Rule:

• Differentiate numerator: $\frac{d}{dx}(2e^{2x}-2) = 4e^{2x}$.
• Differentiate denominator: $\frac{d}{dx}(2x) = 2$.

The limit becomes $\lim_{x\to 0} \frac{4e^{2x}}{2}$. Now, substitute $x=0$: $\frac{4e^{2(0)}}{2} = \frac{4e^0}{2} = \frac{4(1)}{2} = 2$.

Alternatively, using Taylor series expansion for $e^{2x}$ around $x=0$:

$e^{2x} = 1 + (2x) + \frac{(2x)^2}{2!} + \frac{(2x)^3}{3!} + \dots = 1 + 2x + \frac{4x^2}{2} + \frac{8x^3}{6} + \dots = 1 + 2x + 2x^2 + \frac{4}{3}x^3 + \dots$

Substitute this into the expression for $f(x)$:

$f(x) = \frac{(1 + 2x + 2x^2 + \frac{4}{3}x^3 + \dots) - 1 - 2x}{x^2}$

$f(x) = \frac{2x^2 + \frac{4}{3}x^3 + \dots}{x^2}$

Factor out $x^2$ from the numerator:

$f(x) = \frac{x^2(2 + \frac{4}{3}x + \dots)}{x^2}$

Cancel $x^2$ (since $x \neq 0$ as $x \to 0$):

$f(x) = 2 + \frac{4}{3}x + \dots$

Now, take the limit as $x \to 0$:

$\lim_{x\to 0} f(x) = \lim_{x\to 0} (2 + \frac{4}{3}x + \dots) = 2$.