InterviewSolution
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InterviewSolution
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Find the horizontal, vertical and oblique asymptotes of each of the curves. `{:((a),y=x/(x+4),,(b),y=(x^(2)+4)/(x^(2)-1)),((c),y=x^(3)/(x^(2)+3x-10),,(d),y=(x^(3)+1)/(x^(3)+x)),((e),y=x/(root(4)(x^(4)+1)),,(f),y=(x-9)/(sqrt(4x^(2)+3x+2))),((g),y=1/(2^(x)-1),,(h),y=1/(log_(e) x)),((i),y= 1/(2^(x) - 1),,,):}` |
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Answer» (a) ` y= x/(x+4)` Since the degree of numerator and denominator is same, there is a horizontal asymptote y = 1 as `underset(x to infty )"lim"x/(x+4) = underset(x to infty)"lim" 1/(1+4/x) = 1`. Comparing the denominator to zero, we have x+4 = 0 or x =- 4, which is the vertical asymptote. Clearly, the curve has no oblique asymptote. (b) ` y = (x^(2)+4)/(x^(2)-1)` Since the degree of numerator and denominator is same, there is a horizontal asymptote y = 1 as ` underset(x to infty)"lim" (x^(2)+4)/(x^(2)-1) = underset( x to infty)"lim"(1+4/x^(2))/(1-1/x^(2)) = 1`. Comparing denominator to zero, we have ` x^(2) - 1 = 0`. So x = -1 , which are vertical asymptotes. Clearly, the curve has no obluque asymptote. (c) ` y = x^(3)/(x^(2)+3x-10)` Since the degree of numerator is higher than the degree of denominator, three is no horizontal asymptote. Comparing the denominator to zero, we have ` x^(2) + 3x - 10 = 0`. So x = 2 or x = -5, which are vertical asymptotes Also ` y = x- 3+(19x - 30)/(x^(2) + 3x -10)` , so y = x - 3 is an oblique asymptote. (d) ` y = (x^(3)+1)/(x^(3)+x)` Since the degree of numerator is same as the degree of denominator, there is a horizontal asymptote y = 1 as ` underset(x to infty)"lim" (x^(3)+1)/(x^(3)+x)=underset(x to infty)"lim" (1+1/x^(3))/(1-1/x^(2)) = 1`. Comparing the denominator to zero, we have ` x^(3) + x = 0`. So x = 1 , which is a vertical asymptote. Clearly, the curve has no oblique asymptote. (e) ` y = underset(x to infty)"lim" x/root(4)(x^(4)+1)` `underset(x to infty)"lim" 1/(1/xroot(4)(x^(4)+1))=underset(x to infty)"lim" 1/root(4)(1+1/x^(4))= 1` Hence x = 1 is a horizontal asymptote. Similarly, ` underset(x to - infty)"lim" 1/(1/x root(4)(x^(4)+1) )= underset(x to - infty)"lim"1/root(-4)(1+1/x^(4))=-1` So the curve has horizontal asymptotes ` y = pm 1/2`. Clearly, the curve has no vertical as well as oblique asymptote. (g) ` y = sqrt(4x^(2)+3)/(x-2)` ` underset(x to infty)"lim" sqrt(4x^(2)+3)/(x-2) = 2 and underset(x to - infty)"lim" sqrt(4x^(2)+3)/(x-2) =- 2` So the curve has horizontal asymptotes ` y = pm 2`. Comparing the denominator to zero, we have x - 2 = 0 or x = 2, which is a vertical asymptote. (h) ` y = 1/(log_(e)x)` ` underset(x to infty)"lim" 1/(log_(e)x) =0` So y = 0 is a horizontal asymptote. ` underset(x to 1^(+))"lim" 1/(log_(e) x) = infty and underset(x to 1^(-))"lim" 1/(log_(e) x) =- infty` So x = 0 is a vertical asymptote. (i) ` y = 1/(2^(x)-1)` ` underset(x to 0^(+))"lim" 1/(2^(x)-1)= infty underset(x to 0^(-))"lim" 1/(2^(x)-1) =- infty` So x = 0 is a vertical asymptote. ` underset(x to infty)"lim" 1/(2^(x)-1) = 0` So y = 0 is a horizontal asymptote. ` underset(x to - infty)"lim" 1/(2^(x)-1) =- 1` So y =- 1 is a horizontal asymptote. |
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