Verified Answer :

The answer is not infinity. 
Consider FBDs of each mass with the direction of motion of m1 as positive 

m1g-T=m1a 

T-m2g=m2a 

assuming mass less cord and mass less, friction less pulley 
the accelerations are equal 
a=(T-m2g)/m2 

m1g-T=m1(T-m2g)/m2 
do some algebra 
m1g-T=m1T/m2-m1g 
2m1g=T(1+m1/m2) 
2m1m2g=T(m2+m1) 
2m1m2g/(m2+m1)=T 
now take the limit of T as m1->infinity 
T=2m2g 

this is intuitively correct since the maximum acceleration of m1 is -g, the cord transfers the acceleration to m2, which is being acted on by gravity downward and an upward acceleration of g. Therefore the maximum acceleration of m1 is 2g upward. 

batolisis

The value that the magnitude of the tension approach is ; \(T=2*m_{2} *g\)

plagiarism

Assuming  FBDs of masses m₁ and m₂ with the direction of motion of m₁ been positive

Resolving using the relation below

=   \(m_{1} *g-T= m_{1} *g\)

Given that sum of forces on the rope =  

therefore as m₁ → ∞

The magnitude of the tension approach will be :

\(T=2*m_{2} *g\).

Hence we can conclude that The value that the magnitude of the tension approaches is ; \(T=2*m_{2} *g\)

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