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How do you find the limit of #(3x +9) /sqrt (2x^2 +1)# as x approaches infinity?

Answer 1

#3/sqrt2#

Factor the largest-degreed term from the numerator and denominator.

#lim_(xrarroo)(3x+9)/sqrt(2x^2+1)=lim_(xrarroo)(x(3+9/x))/sqrt(x^2(2+1/x^2))#

Pulling the #x^2# out of the square root as just #x# (because we are only concerned with positive value of #x#), this becomes

#lim_(xrarroo)(x(3+9/x))/(xsqrt(2+1/x^2))=lim_(xrarroo)(3+9/x)/sqrt(2+1/x^2)#

Now, when we evaluate the limit as #x# approaches infinity, we see that #9/x# and #1/x^2# both go to #0#.

#=(3+0)/sqrt(2+0)=3/sqrt2#

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Answer 2

Emma Aldridge

To find the limit of (3x + 9) / sqrt(2x^2 + 1) as x approaches infinity, we can simplify the expression by dividing both the numerator and denominator by x. This gives us (3 + 9/x) / sqrt(2 + 1/x^2).

As x approaches infinity, the term 9/x approaches 0, and the term 1/x^2 also approaches 0. Therefore, the expression simplifies to 3 / sqrt(2).

Hence, the limit of (3x + 9) / sqrt(2x^2 + 1) as x approaches infinity is 3 / sqrt(2).

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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.