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 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 #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#
