# 数学代写|微积分代写Calculus代写|MTH2010

## 数学代写|微积分代写Calculus代写|Vertical asymptotes

One way for a limit not to exist is if it renders the real result $\infty$ or $-\infty$.
Example 9 Determine $\lim {x \rightarrow 4} \frac{x+3}{(x-4)^2}$. Solution This is not a one-sided limit (we have $x \rightarrow 4$, not $x \rightarrow 4^{+}$ or $x \rightarrow 4^{-}$). As mentioned, such limits are sometimes called two-sided limits. We therefore use $x=4+\alpha$ : \begin{aligned} \lim {x \rightarrow 4} \frac{x+3}{(x-4)^2} &=\frac{4+\alpha+3}{(4+\alpha-4)^2}=\frac{7+\alpha}{\alpha^2} \ & \approx \frac{7}{\alpha^2}=7 A^2 \ & \doteq \infty \end{aligned}
We are not stuck, because squared values cannot be negative; we know that $7 A^2$ is positive. Because the limit is infinite, the limit does not exist.

What is the graphical interpretation of this limit? Because the limit’s value is $\infty$, we know that near $x=4$, the values of the expression must become infinite. Therefore, the graph must rise above a $y$-coordinate of 100 , above 1000 , above 1000000 , above any real number! The function’s values must therefore rise above the top of our picture. We express this idea by saying that the function has a vertical asymptote at $x=4$. We denote the vertical asymptote in a graph by a dotted line. The graph of this function is in figure 12 .

In example 2, division by zero resulted in a hole in the graph. In example 9 , division by zero resulted in a vertical asymptote. Merely knowing that we are dividing by zero does not tell us what is happening on the graph; we can only tell by determining the limit.

## 数学代写|微积分代写Calculus代写|Limit examples with piecewise-defined functions

Sometimes a piecewise-defined function can be evaluated as simply as any other limit.
Example 1 Let $f(x)=\left{\begin{array}{ll}2 x, & x \leq 0 \ 4 x, & x>0\end{array}\right.$. Investigate $\lim {x \rightarrow-3} f(x)$. Solution Because the limit is two-sided, we use $x=-3+\alpha$ : $$\lim {x \rightarrow-3} f(x)=f(-3+\alpha)=2(-3+\alpha)=-6+2 \alpha \approx-6 .$$
The limit is $-6$.
Recall that when evaluating a piecewise-defined function, the first task is to determine which rule applies to the function’s input. This was not a difficult task in example 1 . If, however, the limit is requested at a value where the function rules change, the situation can be a little more complicated.
Example 2 Let $f(x)=\left{\begin{array}{cl}x, & x \leq 0 \ x^2, & x>0\end{array}\right.$. Investigate $\lim {x \rightarrow 0} f(x)$. Solution Because the limit is two-sided, we use $x=0+\alpha=\alpha$ : $$\lim {x \rightarrow 0} f(x)=f(\alpha),$$
and we are stuck because we do not know which rule to use to evaluate $f$. If $\alpha>0$, then we use the rule $f(x)=x^2$; but, if $\alpha<0$, we use the rule $f(x)=x$. Just as when we are stuck for other reasons, let’s use one-sided limits to help. We evaluate the limit from the right by using $x=0+\omega=\omega$
$$\lim _{x \rightarrow 0^{+}} f(x)=f(\omega)=\omega^2 \doteq 0 .$$

We are not stuck because $\omega$ is positive and we can use the rule for $x>0$, which is $f(x)=x^2$. The limit from the right is 0 .
We evaluate the limit from the left using $x=0-\omega=-\omega$ :
$$\lim _{x \rightarrow 0^{-}} f(x)=f(-\omega)=-\omega \doteq 0 .$$
We are not stuck because $-\omega$ is negative and we can use the rule for $x \leq 0$, which is $f(x)=x$. The limit from the left is 0 .

Because both one-sided limits have the same value, 0 , the two-sided limit exists:
$$\lim _{x \rightarrow 0} f(x)=0$$

# 微积分代考

## 数学代写|微积分代写Calculus代写|Vertical asymptotes

$$\lim x \rightarrow 4 \frac{x+3}{(x-4)^2}=\frac{4+\alpha+3}{(4+\alpha-4)^2}=\frac{7+\alpha}{\alpha^2} \quad \approx \frac{7}{\alpha^2}=7 A^2 \doteq \infty$$

## 数学代写|微积分代写Calculus代写|Limit examples with piecewise-defined functions

$$我们没有被困住，因为 -\omega 是负数，我们可以使用规则 x \leq 0 ，即 f(x)=x. 左边的限制是 0 。 因为两个单边限制具有相同的值 0 ，所以存在双边限制:$$
\lim _{x \rightarrow 0} f(x)=0


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