How do you evaluate, using infinite limits, for the function y=x/(sqrt(x^2+1)) so as to find the asymptotes?

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The function `y = x/(sqrt(x^2+1))` only has horizontal asymptotes.

As x tends to inf.

`lim_(x->oo)x/sqrt(x^2+1)`

as x tend to inf., 1/x tends to 0

=> `lim_(1/x->0)x/(x*sqrt(1+1/x^2))`

=> `lim_(1/x->0)1/sqrt(1+1/x^2)`

substitute 1/x = 0

=> 1

As x tends to -inf.

`lim_(x->-oo)x/sqrt(x^2+1)`

=> `lim_(x->oo) (-x)/sqrt((-x)^2+1)`

=> `lim_(x->oo)(-x)/sqrt(x^2+1)`

As x tends to inf.,...

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The function `y = x/(sqrt(x^2+1))` only has horizontal asymptotes.

As x tends to inf.

`lim_(x->oo)x/sqrt(x^2+1)`

as x tend to inf., 1/x tends to 0

=> `lim_(1/x->0)x/(x*sqrt(1+1/x^2))`

=> `lim_(1/x->0)1/sqrt(1+1/x^2)`

substitute 1/x = 0

=> 1

As x tends to -inf.

`lim_(x->-oo)x/sqrt(x^2+1)`

=> `lim_(x->oo) (-x)/sqrt((-x)^2+1)`

=> `lim_(x->oo)(-x)/sqrt(x^2+1)`

As x tends to inf., 1/x tends to 0

=> `lim_(1/x->0)(-x)/(x*sqrt(1+1/x^2))`

=> `lim_(1/x->0)(-1)/sqrt(1+1/x^2)`

substitute 1/x = 0

=> -1

The asymptotes are y = 1 and y = -1

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