SigmaJay From: US Posts: 5 Votes: 10 |
This is a Lim as h moves to zero question. I have it simplified to this so far: [√(h+1) + 1]/h I know it's not exactly a calculus question in its simplification but the "as h moves to zero" is. Please show me the steps! NO THE ANSWER IS NOT INFINITY. it is 1/2, that's the slope of the line at a given point. I looked up the answer but i need to know how to get there. |

TinyMath9 From: US Posts: 20 Votes: 60 |
Lim h->0 [√(h+1) - 1]/h *[√(h+1) + 1]/[√(h+1) + 1] = lim h->0 {h / h* [√(h+1) + 1]} = lim h->0 [1/ (√(h+1) + 1)] = 1/ (√(0+1) + 1) = 1/2 |

WizNewton From: US Posts: 26 Votes: 52 |
The equation you have described doesn't cross the y axis at all. h cannot equal zero. If however the equation is [√(h+1) - 1]/h, then both the top and bottom tend to zero, and you need to use l'Hopital's rule and differentiate the top and the bottom to come up with a reasonable answer. differentiating top: (1/2)/√(h+1) differentiation bottom: 1 so the answer is the limit as h->0 of 1/2*√(h+1), which is a half. |

BabyBrainy4 From: US Posts: 29 Votes: 87 |
You've got a little bit of an error in your equation if you're looking for a slope. It should be (sqrt(h+1) MINUS 1)/h Simplifies to: sqrt(1/h + 1/h^2) - 1/h multiply the denominator and the numerator by sqrt(1/h + 1/h^2) + 1/h (1/h +(1/h^2 - 1/h^2))/(sqrt(1/h + 1/h^2) + 1/h) multiply top and bottom by h 1/(sqrt(h + 1) + 1) evaluate the limit 1/2 Hope that helps :) |

ProSam1 From: US Posts: 6 Votes: 18 |
Lim h→0 of [√(h+1) + 1]/h Rationalize numerator by multiplying both numerator and denominator by [√(h+1) − 1] we get lim h→0 of [√(h+1) + 1]/h =lim h→0 of [{√(h+1) + 1×}{√(h+1) − 1]}]/[h×{√(h+1) − 1}] =lim h→0 of [{√(h+1)}² − 1}]/[h×{√(h+1) − 1}] =lim h→0 of [(h+1) − 1}]/[h×{√(h+1) − 1}] =lim h→0 of (h}/[h×{√(h+1) − 1}] =lim h→0 of 1/{√(h+1) − 1} =1/{√(0+1) − 1}=1/{1 − 1} =1/0 i.e limit does not exist. I believe your simplification is wrong Because lim h→0 of [√(h+1) − 1]/h is 1/2 Pl check. |

Anonymous From: - Posts: - Votes: - |
I can't help feeling that original equation must be f(x) = √x and you are asked to find f'(1), which will give: f'(1) = lim[h→0] (f(1+h) - f(1)) / h . . . . = lim[h→0] (√(1+h) - √1) / h . . . . = lim[h→0] (√(1+h) - 1) / h . . . . = lim[h→0] (√(1+h) - 1)(√(1+h) + 1) / (h(√(1+h) + 1)) . . . . = lim[h→0] ((1+h) - 1) / (h(√(1+h) + 1)) . . . . = lim[h→0] h / (h(√(1+h) + 1)) . . . . = lim[h→0] 1 / (√(1+h) + 1) . . . . = 1 / (√(1+0) + 1) . . . . = 1 / (1 + 1) . . . . = 1 / 2 NOTE: the expression you gave us [√(h+1) + 1]/h will give a limit of ∞ as h approaches 0 |

Anonymous From: - Posts: - Votes: - |
Are you sure it is +1 and not -1? You can expand (h+1)^0.5 as 1 + h/2 + 1/2.-1/2 h^2/2! + ......... - if the number was -1 then this would simplify to [h/2 + .... ]/h which gives you the answer you expect |

ProfessorMath3 From: US Posts: 44 Votes: 44 |
The limit will have a finite value only if you take the negative root of (h + 1) and the answer is -1/2 i can convert it into derivative form and solve it, but not sure how to proceed directly |

Anonymous From: - Posts: - Votes: - |
The question u ve given seems to be wrong, check it if it is +1 or -1Lim h->0 [√(h+1) - 1]/h *[√(h+1) + 1]/[√(h+1) + 1] = lim h->0 {h / h* [√(h+1) + 1]} = lim h->0 [1/ (√(h+1) + 1)] = 1/ (√(0+1) + 1) = 1/2 |

Anonymous From: - Posts: - Votes: - |
Bvmnmvm |

Anonymous From: - Posts: - Votes: - |
Bvmnmvm |

Anonymous From: - Posts: - Votes: - |
It is good iam empressive |

Anonymous From: - Posts: - Votes: - |
It is good iam empressive |

Anonymous From: - Posts: - Votes: - |
OK LOL |

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