Given expression
#lim_(x→0)(e^(x^2 )+e^(〖-x〗^2 )-2)/x^2?#
We see that the expression becomes #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 #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#
