Verified answer

Question - 1:

i) Dividing both numerator and denominator by x, [In such case, since numerator being a square root function, inside square root it is divided by x² ]

√(x² - 9)/(2x - 6) = √(1 - 9/x²)/(2 - 6/x)

ii) As x ->∞, both 9/x² and 6/x --> 0

So limit of the function = √(1 - 0)/(2 - 0) = 1/2

Question - 2:

ii) Rationalizing the numerator,

The given function = [(x² + 4x + 1) - (x²)]/[√(x² + 4x + 1) + x]

This simplifies to: (4x + 1)/[√(x² + 4x + 1) + x]

Now dividing by x both numerator and denominator as in previous problem,

The given function = (4 + 1/x)/√(1 + 4/x + 1/x²) + 1]

As x tends to infinity, 1/x, 4/x and 1/x² tends to zero.

Thus limit of the function = (4 + 0)/[√(1 + 0 + 0) + 1] = 4/2 = 2

Lim x-->0 sin(-3x)/4x considering the two applications are non-provide up at x = 0 and the cost of the branch at x = 0 is 0/0. Then we are able to apply h’Lopital’s Rule: Lim x-->0 sin(-3x)/4x = Lim x-->0 -3cos(-3x)/4 = -3/4

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