The value of
x→0lim(3−9+x5x)
is ....
Explanation
To solve the limit limx→0(3−9+x5x), we rationalize the denominator
Rationalizing the Denominator
Multiply both numerator and denominator by the conjugate of the denominator, which is 3+9+x
x→0lim(3−9+x5x)=x→0lim(3−9+x5x⋅3+9+x3+9+x)
=x→0lim((3−9+x)(3+9+x)5x(3+9+x))
Simplifying the Denominator
Use the identity (a−b)(a+b)=a2−b2
=x→0lim(9−(9+x)5x(3+9+x))
=x→0lim(9−9−x5x(3+9+x))
=x→0lim(−x5x(3+9+x))
Simplifying
Simplify by dividing x in the numerator and denominator
=x→0lim−xx(15+59+x)
=x→0lim(−15−59+x)
Substitute x=0
=−15−59+0
=−15−5⋅3
=−15−15=−30
Therefore, the value of limx→0(3−9+x5x)=−30