Last Activity: 13 Years ago
the above function takes ∞-∞ indeterminant form
rationalising it we get
lim x→∞ [√(2x2-3)-√(2) x]*[√(2x2-3)+√(2) x]/[√(2x2-3)+√(2) x]
=lim x→∞ -3/[√(2x2-3)+√(2) x]
this is not an indeterminant form so put x=∞ in the expression
=-3/(∞+∞)
=0
hope this helps
Hi Menka,
lim x→∞ x[√(2-3/x2) - √2]
lim x→∞ x [(2-3/x2 - 2)/(√(2-3/x2)+√2)] ------------- {Multiplying Nr and Dr by the conjugate of the Nr}
lim x→∞ -3/[(x)*(√(2-3/x2)+√2)].
Now this, as x→∞, will tend to 0. {As x is in the Dr}.
Hence the limit of the question is 0.
Hope that helps.
All the Best,
Regards,
Ashwin.
lim x->inf (2x^2-3)^0.5-2^0.5x
Replace x by 1/y such that y ->0
so lim y->0 (1/y){(2-3y^2)^0.5-2^0.5}[0/0 form]
=lim y->0 (-3)y/(2-3y^2)^0.5[L Hospital's rule] =0
The answer is WRONG.
we never discuss ∞+∞ forms.
The answer is Wrong for 2 reasons
1) The given limit is in the form ∞-∞ which cannot be solved
2) Look for the image below
Hi Math Solver,
Kindly note that the denominator as x→∞, will be of the form ∞*(√2+√2).....
Qnd hence the limiting value is zero....
Best Regards,
Ashwin (IIT Madras).
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