# 請教3題函數與極限

## 請教3題函數與極限

103.3.18幫忙輸入題目
1.設$$\displaystyle y=\sum_{k=1}^{20} |\;x-k|\;+3x+2$$，則y的最小值?此時x之值?

2.$$\displaystyle \lim_{x \to 1} \frac{\sqrt{2x^2+a}-x+b}{(x-1)^2}=$$定值,求a,b?

3.設$$a,b,x \in R$$,若$$\displaystyle f(x)=\lim_{n \to \infty}\frac{x^{2n-1}+ax+b}{x^{2n}+1}$$為連續函數,則求a,b?

[ 本帖最後由 bugmens 於 2014-3-18 04:15 PM 編輯 ]

#### 附件

1.pdf (50.02 KB)

2014-3-18 15:57, 下載次數: 1999

TOP

## 回復 1# thankyou 的帖子

$$\displaystyle \Rightarrow \sqrt{2+a}-1+b=0\Rightarrow a=b^2-2b-1$$

$$\displaystyle \Rightarrow \lim_{x\to1}\frac{\sqrt{2x^2+a}-x+b}{\left(x-1\right)^2}=\lim_{x\to1}\frac{\sqrt{2x^2+b^2-2b-1}-x+b}{\left(x-1\right)^2}$$

$$\displaystyle =\lim_{x\to1}\frac{2x^2+b^2-2b-1-\left(x-b\right)^2}{\left(x-1\right)^2\left(\sqrt{2x^2+b^2-2b-1}+x-b\right)}$$

$$\displaystyle =\lim_{x\to1}\frac{x^2-1+2b\left(x-1\right)}{\left(x-1\right)^2\left(\sqrt{2x^2-b-1}+x-b\right)}$$

$$\displaystyle =\lim_{x\to1}\frac{\left(x+1+2b\right)\left(x-1\right)}{\left(x-1\right)^2\left(\sqrt{2x^2-b-1}+x-b\right)}$$

$$\displaystyle =\lim_{x\to1}\frac{x+1+2b}{\left(x-1\right)\left(\sqrt{2x^2-b-1}+x-b\right)}$$

$$\Rightarrow 1+1+2b=0\Rightarrow b=-1, a=b^2-2b-1=2$$

$$\displaystyle \Rightarrow \lim_{x\to1^{-}}f(x)=a+b$$ 且 $$\displaystyle \lim_{x\to-1^{+}}f(x)=-a+b$$

$$\displaystyle \Rightarrow \lim_{x\to1^{+}}f(x)=1$$ 且 $$\displaystyle \lim_{x\to-1^{-}}f(x)=-1$$

$$\displaystyle \Rightarrow 1=a+b=\frac{1+a+b}{2}$$ 且 $$\displaystyle -a+b=-1=\frac{-1-a+b}{2}$$

TOP

TOP

﻿