# Question fda26

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 different 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 )+(-2x)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)(-2x)))/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#