数学代写|微积分代写Calculus代写|MATH141

数学代写|微积分代写Calculus代写|Evaluating Limits by Direct Substitution

In Example 2(a) we determined that $\lim _{x \rightarrow 5} f(x)=39$, where $f(x)=2 x^2-3 x+4$. Notice that $f(5)=39$; in other words, we would have gotten the correct result simply by substituting 5 for $x$. Similarly, direct substitution provides the correct answer in part (b). The functions in Example 2 are a polynomial and a rational function, respectively, and similar use of the Limit Laws proves that direct substitution always works for such functions (see Exercises 59 and 60). We state this fact as follows.

Functions that have the Direct Substitution Property are called continuous at $a$ and will be studied in Section 2.5. However, not all limits can be evaluated initially by direct substitution, as the following examples show.
EXAMPLE 3 Find $\lim {x \rightarrow 1} \frac{x^2-1}{x-1}$ SOLUTION Let $f(x)=\left(x^2-1\right) /(x-1)$. We can’t find the limit by substituting $x=1$ because $f(1)$ isn’t defined. Nor can we apply the Quotient Law, because the limit of the denominator is 0 . Instead, we need to do some preliminary algebra. We factor the numerator as a difference of squares: $$\frac{x^2-1}{x-1}=\frac{(x-1)(x+1)}{x-1}$$ The numerator and denominator have a common factor of $x-1$. When we take the limit as $x$ approaches 1 , we have $x \neq 1$ and so $x-1 \neq 0$. Therefore we can cancel the common factor, $x-1$, and then compute the limit by direct substitution as follows: \begin{aligned} \lim {x \rightarrow 1} \frac{x^2-1}{x-1} &=\lim {x \rightarrow 1} \frac{(x-1)(x+1)}{x-1} \ &=\lim {x \rightarrow 1}(x+1)=1+1=2 \end{aligned}
The limit in this example arose in Example 2.1.1 in finding the tangent to the parabola $y=x^2$ at the point $(1,1)$.

NOTE In Example 3 we were able to compute the limit by replacing the given function $f(x)=\left(x^2-1\right) /(x-1)$ by a simpler function, $g(x)=x+1$, that has the same limit. This is valid because $f(x)=g(x)$ except when $x=1$, and in computing a limit as $x$ approaches 1 we don’t consider what happens when $x$ is actually equal to 1 . In general, we have the following useful fact.

数学代写|微积分代写Calculus代写|The Precise Definition of a Limit

To motivate the precise definition of a limit, let’s consider the function
$$f(x)= \begin{cases}2 x-1 & \text { if } x \neq 3 \ 6 & \text { if } x=3\end{cases}$$
Intuitively, it is clear that when $x$ is close to 3 but $x \neq 3$, then $f(x)$ is close to 5 , and so $\lim _{x \rightarrow 3} f(x)=5$.

To obtain more detailed information about how $f(x)$ varies when $x$ is close to 3 , we ask the following question:
How close to 3 does $x$ have to be so that $f(x)$ differs from 5 by less than $0.1$ ?
The distance from $x$ to 3 is $|x-3|$ and the distance from $f(x)$ to 5 is $|f(x)-5|$, so our problem is to find a number $\delta$ (the Greek letter delta) such that
$$|f(x)-5|<0.1 \quad \text { if } \quad|x-3|<\delta \quad \text { but } x \neq 3$$ If $|x-3|>0$, then $x \neq 3$, so an equivalent formulation of our problem is to find a number $\delta$ such that
$$|f(x)-5|<0.1 \quad \text { if } \quad 0<|x-3|<\delta$$
Notice that if $0<|x-3|<(0.1) / 2=0.05$, then
$$|f(x)-5|=|(2 x-1)-5|=|2 x-6|=2|x-3|<2(0.05)=0.1$$
that is, $\quad|f(x)-5|<0.1 \quad$ if $\quad 0<|x-3|<0.05$
Thus an answer to the problem is given by $\delta=0.05$; that is, if $x$ is within a distance of $0.05$ from 3 , then $f(x)$ will be within a distance of $0.1$ from 5 .

If we change the number $0.1$ in our problem to the smaller number $0.01$, then by using the same method we find that $f(x)$ will differ from 5 by less than $0.01$ provided that $x$ differs from 3 by less than $(0.01) / 2=0.005$ :
$$|f(x)-5|<0.01 \quad \text { if } \quad 0<|x-3|<0.005$$
Similarly,
$$|f(x)-5|<0.001 \quad \text { if } \quad 0<|x-3|<0.0005$$

微积分代考

数学代写|微积分代写Calculus代写|Evaluating Limits by Direct Substitution

$$\frac{x^2-1}{x-1}=\frac{(x-1)(x+1)}{x-1}$$

$$\lim x \rightarrow 1 \frac{x^2-1}{x-1}=\lim x \rightarrow 1 \frac{(x-1)(x+1)}{x-1} \quad=\lim x \rightarrow 1(x+1)=1+1=2$$

$f(x)=\left(x^2-1\right) /(x-1)$ 通过一个更简单的函数， $g(x)=x+1$ ，具有相同的极 限。这是有效的，因为 $f(x)=g(x)$ 除了什么时候 $x=1$ ，并且在计算极限时 $x$ 方法 1 我们不考虑什么时候发生 $x$ 实际上等于 1。总的来说，我们有以下有用的事实。

数学代写|微积分代写Calculus代写|The Precise Definition of a Limit

$$f(x)={2 x-1 \quad \text { if } x \neq 36 \quad \text { if } x=3$$

$$|f(x)-5|<0.1 \quad \text { if } \quad|x-3|<\delta \quad \text { but } x \neq 3$$ 如果 $|x-3|>0$ ， 然后 $x \neq 3$ ，所以我们的问题的等效表述是找到一个数字 $\delta$ 这样
$$|f(x)-5|<0.1 \quad \text { if } \quad 0<|x-3|<\delta$$

$$|f(x)-5|=|(2 x-1)-5|=|2 x-6|=2|x-3|<2(0.05)=0.1$$

$$|f(x)-5|<0.01 \quad \text { if } \quad 0<|x-3|<0.005$$

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