Use series to evaluate the limit.
-$$\lim _ { x \rightarrow \infty } x ^ { 2 } \left( e ^ { 4 / x ^ { 2 } } - 1 \right)$$
A) -1
B) 1
C) 4
D) -4

Question

Quizplus · Accepted Answer

First, we use the Maclaurin series expansion of $e^x$:
$$e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \dotsb$$
Substituting $x = \frac{4}{x^2}$ gives:
$$e^{\frac{4}{x^2}} = 1 + \frac{4}{x^2} + \frac{2^2}{2! x^4} + \frac{2^3}{3! x^6} + \dotsb$$
Multiplying by $x^2$ and subtracting 1 gives:
$$x^2 \left( e ^ { 4 / x ^ { 2 } } - 1 ight) = 4 + \frac{2^2}{1! x^2} + \frac{2^3}{2! x^4} + \frac{2^4}{3! x^6} + \dotsb$$
As $x 	o \infty$, every term on the right-hand side goes to zero except for the first term $4$. Therefore, the limit is:
$$\lim_{x 	o \infty} x^2 \left( e ^ { 4 / x ^ { 2 } } - 1 ight) = \lim_{x 	o \infty} 4 = \boxed{4}$$

Use Series to Evaluate the Limit lim⁡x→∞x2(e4/x2−1)\lim _ { x \rightarrow \infty } x ^ { 2 } \left( e ^ { 4 / x ^ { 2 } } - 1 \right)x→∞lim​x2(e4/x2−1)

Use Series to Evaluate the Limit $\lim _ { x \rightarrow \infty } x ^ { 2 } \left( e ^ { 4 / x ^ { 2 } } - 1 \right)$