How do you find #lim (x+x^(1/2)+x^(1/3))/(x^(2/3)+x^(1/4))# as #x->oo# using l'Hospital's Rule?

Answer 1

#lim_(x->oo) (x+x^(1/2)+x^(1/3))/(x^(2/3)+x^(1/4)) = +oo#

You do not really need l'Hospital's rule as:

#(x+x^(1/2)+x^(1/3))/(x^(2/3)+x^(1/4)) = (x^(2/3) (x^(1-2/3) +x^(1/2-2/3) +x ^(1/3-2/3))) / (x^(2/3)( 1 + x^(1/4-2/3))#
#(x+x^(1/2)+x^(1/3))/(x^(2/3)+x^(1/4)) = (cancel(x^(2/3)) (x^(1/3) +x^(-1/6) +x ^(-1/3))) / (cancel(x^(2/3))( 1 + x^(-5/12))#
#(x+x^(1/2)+x^(1/3))/(x^(2/3)+x^(1/4)) = (x^(1/3) +x^(-1/6) +x ^(-1/3)) / ( 1 + x^(-5/12))#

so as:

#lim_(x->oo) x^(1/3) +x^(-1/6) +x ^(-1/3) = +oo#
#lim_(x->oo) 1 + x^(-5/12) = 1#
#lim_(x->oo) (x+x^(1/2)+x^(1/3))/(x^(2/3)+x^(1/4)) = +oo#
Anyway, as the limit is in the form #oo/oo# we can apply l'Hospital's rule:
#lim_(x->oo) (x+x^(1/2)+x^(1/3))/(x^(2/3)+x^(1/4)) = lim_(x->oo) (d/dx (x+x^(1/2)+x^(1/3)))/(d/dx (x^(2/3)+x^(1/4)) )#
#lim_(x->oo) (x+x^(1/2)+x^(1/3))/(x^(2/3)+x^(1/4))= lim_(x->oo) (1+1/2x^(-1/2)+1/3x^(-1/3))/(2/3x^(-1/3)+1/4x^(-3/4)) = 1/0 = +oo#
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Answer 2

To find lim (x+x^(1/2)+x^(1/3))/(x^(2/3)+x^(1/4)) as x approaches infinity using l'Hopital's Rule, we proceed as follows:

  1. Take the derivative of the numerator and denominator separately.
  2. Evaluate the limit of the resulting expressions as x approaches infinity.
  3. If the limit remains indeterminate (e.g., 0/0 or ∞/∞), repeat the process until a determinate form is obtained.

Let's apply l'Hopital's Rule:

First, find the derivatives:

Numerator: f'(x) = 1 + (1/2)x^(-1/2) + (1/3)x^(-2/3)

Denominator: g'(x) = (2/3)x^(-1/3) + (1/4)x^(-3/4)

Now, take the limit of the derivatives as x approaches infinity:

lim (x+x^(1/2)+x^(1/3))/(x^(2/3)+x^(1/4)) as x approaches infinity = lim (1 + (1/2)x^(-1/2) + (1/3)x^(-2/3)) / ((2/3)x^(-1/3) + (1/4)x^(-3/4)) as x approaches infinity

= lim ((1/2)x^(-1/2) + (1/3)x^(-2/3)) / ((2/3)x^(-1/3) + (1/4)x^(-3/4)) as x approaches infinity

Now, apply l'Hopital's Rule again:

= lim ((-1/4)x^(-3/2) - (2/9)x^(-5/3)) / ((-2/9)x^(-4/3) - (3/16)x^(-7/4)) as x approaches infinity

Since the degree of the numerator and denominator are the same, we can evaluate the limit by dividing the leading coefficients:

= (-1/4) / (-2/9) as x approaches infinity = 9/8

Therefore, lim (x+x^(1/2)+x^(1/3))/(x^(2/3)+x^(1/4)) as x approaches infinity equals 9/8.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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