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How do you find the limit of #x / sqrt (4x^2 + 2x +1) # as x approaches infinity?

Answer 1

Cameron Alexander

Please see below.

#x / sqrt (4x^2 + 2x +1) = x / (sqrt(x^2)sqrt (4 + 2/x +1/x^2))# #" "# for #x != 0#

#sqrt(x^2) = absx = x# for #x > 0#

#lim_(xrarroo) x/ sqrt (4x^2 + 2x +1) = lim_(xrarroo)x / (xsqrt (4 + 2/x +1/x^2))#

# = lim_(xrarroo) 1/ (sqrt (4 + 2/x +1/x^2))#

# = 1/sqrt(4+0+0) = 1/2#

Bonus material

Since #sqrt(x^2) = -x# for #x < 0#, we have

#lim_(xrarr-oo) x/ sqrt (4x^2 + 2x +1) = lim_(xrarr-oo)x / (-xsqrt (4 + 2/x +1/x^2))#

# = lim_(xrarr-oo) 1/ (-sqrt (4 + 2/x +1/x^2))#

# = -1/sqrt(4+0+0) = - 1/2#

Here is the graph

graph{x / sqrt (4x^2 + 2x +1) [-10, 10, -5, 5]}

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Answer 2

Christopher Agresta

To find the limit of x / sqrt(4x^2 + 2x + 1) as x approaches infinity, we can simplify the expression by dividing both the numerator and denominator by x. This gives us 1 / sqrt(4 + 2/x + 1/x^2). As x approaches infinity, both 2/x and 1/x^2 approach zero. Therefore, the expression simplifies to 1 / sqrt(4) = 1/2. Thus, the limit of x / sqrt(4x^2 + 2x + 1) as x approaches infinity is 1/2.

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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.