A) This particular limit is of the indeterminate form,

if we plug in infinity directly, though it is not a number just to check.
If a limit is in this form, we apply L'Hopital's Rule.
's

So we take the derivatives and obtain,
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Still it is of the same indeterminate form, so we apply the rule again,

This simplifies to,
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b) This limit is also of the indeterminate form,

we still apply the L'Hopital's Rule,


When we plug in zero now we obtain,
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c) This also in the same indeterminate form


It is still of that indeterminate form so we apply the rule again, to obtain;

Now we have remove the discontinuity, we can evaluate the limit now, plugging in zero to obtain;

This gives us;

d)

For this kind of question we need to rationalize the radical function, to obtain;

We now divide both the numerator and denominator by x, to obtain,

This simplifies to,