# Question ac803

Feb 13, 2017

Zero

#### Explanation:

Given expression
lim_(x→0)⁡(e^(x^2 )+e^(〖-x〗^2 )-2)/x^2?
We see that the expression becomes $\frac{0}{0}$ if we calculate its value at $x = 0$. Hence L' Hospitals rule is applicable. We differentiate the numerator and denominator with rest to $x$ and then evaluate.

=>lim_(x→0)⁡ (d/dx[e^(x^2 )+e^([-x]^2 )-2])/(d/dxx^2)
Using the chain rule we get
lim_(x→0)⁡ (2xe^(x^2 )+2(-x)e^([-x]^2 ))/(2x)
=>lim_(x→0)⁡ (xe^(x^2 )-xe^([-x]^2 ))/(x)

When we evaluate the function at $x = 0$ we see that it is $\frac{0}{0.}$ Hence applying L' Hospitals rule again we get
lim_(x→0)⁡ (d/dx[xe^(x^2 )-xe^([-x]^2 )])/(d/dx x)
Using product rule of differentiation we get
lim_(x→0)⁡ ((e^(x^2 )+xe^(x^2)(2x))-(e^([-x]^2 )+xe^([-x]^2)2(-x)))/1
=>lim_(x→0)⁡ ((e^(x^2 )+2x^2e^(x^2))-(e^([-x]^2 )-2x^2e^([-x]^2)))/1
=>lim_(x→0)⁡ (1+0-1+0)/1=0#