作业帮微积分3 找极限的求答案
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微积分3 找极限的求答案
Find the limit of the function f(x; y) =( 2^y + 2^(-y) -2 )/x^2
along the line y=38.3x as x approaches 0.
Find the limit of the function f(x; y) =( 2^y + 2^(-y) -2 )/x^2
along the line y=38.3x as x approaches 0.
The first floor's answer is wright.
lim f(x; y) =lim ( 2^y + 2^(-y) -2 )/x^2
x->0 x->0
y=38.3x y=38.3x
=lim [2^y*ln2*dy/dx + 2^(-y)*ln2*(-dy/dx)]/(2x)=lim [2^y*ln2*38.3 -2^(-y)*ln2*38.3]/(2x)
x->0 x->0
y=38.3x y=38.3x
=lim [2^y*ln2*38.3*ln2*38.3 -2^(-y)*ln2*38.3*ln2*(-38.3)]/2
x->0
y=38.3x
=(38.3ln2)^2=704.7717
lim f(x; y) =lim ( 2^y + 2^(-y) -2 )/x^2
x->0 x->0
y=38.3x y=38.3x
=lim [2^y*ln2*dy/dx + 2^(-y)*ln2*(-dy/dx)]/(2x)=lim [2^y*ln2*38.3 -2^(-y)*ln2*38.3]/(2x)
x->0 x->0
y=38.3x y=38.3x
=lim [2^y*ln2*38.3*ln2*38.3 -2^(-y)*ln2*38.3*ln2*(-38.3)]/2
x->0
y=38.3x
=(38.3ln2)^2=704.7717