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What are the asymptote(s) and hole(s), if any, of # f(x) =(sinx)/(5x^2+2x+1) #?

Answer 1

Please see below.

There are no holes and no vertical asymptotes because the denominator is never #0# (for real #x#).

Using the squeeze theorem at infinity we can see that #lim_(xrarroo)f(x) = 0# and also #lim_(xrarr-oo)f(x) = 0#, so the #x#-axis is a horizontal asymptote.

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Answer 2

Abigail Allen

The function f(x) = (sinx)/(5x^2+2x+1) has no holes. The vertical asymptotes occur when the denominator is equal to zero, which in this case is at x = -0.2 and x = -1. The horizontal asymptote is y = 0.

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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.