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# 数学代写|微积分代写Calculus代考|Miscellaneous algebra

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## 数学代写|微积分代写Calculus代考|Miscellaneous algebra

When factoring and conjugate multiplication don’t work, try other basic algebra like adding or subtracting fractions, multiplying or dividing fractions, canceling, or some other form of simplification.
Evaluate $\lim _{x \rightarrow 0} \frac{\frac{1}{x+4}-\frac{1}{4}}{x}$.

Try substitution.
Plug in 0: That gives you $\frac{0}{0}$ – no good.

Simplify the complex fraction (that’s a big fraction that contains little fractions) by multiplying the numerator and denominator by the least common denominator of the little fractions, namely $4(x+4)$.

Note: Adding or subtracting the little fractions in the numerator or denominator also works in this type of problem, but it’s a bit longer than the method here.
\begin{aligned} & \lim {x \rightarrow 0} \frac{\left(\frac{1}{x+4}-\frac{1}{4}\right)}{x} \cdot \frac{4(x+4)}{4(x+4)} \ = & \lim {x \rightarrow 0} \frac{4-(x+4)}{4 x(x+4)} \ = & \lim {x \rightarrow 0} \frac{-x}{4 x(x+4)} \ = & \lim {x \rightarrow 0} \frac{-1}{4(x+4)} \end{aligned}

1. Now substitution works.
$$=\frac{-1}{4(0+4)}=-\frac{1}{16}$$

## 数学代写|微积分代写Calculus代考|Limits at Infinity

In the limits in the last section, $x$ approaches a finite number, but there are also limits where $x$ approaches infinity or negative infinity. Consider the function $f(x)=\frac{1}{x}$. See Figure 3-1.

You can see on the graph (in the first quadrant) that as $x$ gets bigger and bigger – in other words, as $x$ approaches infinity – the height of the function gets lower and lower but never gets to zero. This is confirmed by considering what happens when you plug bigger and bigger numbers into $\frac{1}{x}$. The outputs get smaller and smaller. This graph thus has a horizontal asymptote of $y=0$ (the $x$-axis), and we say that $\lim {x \rightarrow \infty} \frac{1}{x}=0$. The fact that $x$ never actually reaches infinity and that $f$ never gets to zero has no relevance. When we say that $\lim {x \rightarrow \infty} \frac{1}{x}=0$, we mean that as $x$ gets bigger and bigger without end, $f$ gets closer and closer to zero. The function $f$ also approaches zero as $x$ approaches negative infinity, written as $\lim _{x \rightarrow-\infty} \frac{1}{x}=0$.

Horizontal asymptotes
Horizontal asymptotes and limits at infinity go hand in hand you can’t have one without the other. For a rational function like $f(x)=\frac{3 x-7}{2 x+8}$, determining the limit at infinity or negative infinity is the same as finding the location of the horizontal asymptote.
Here’s what you do. First, note the degree of the numerator (that’s the highest power of $x$ in the numerator) and the degree of the denominator. You’ve got three cases:
If the degree of the numerator is greater than the degree of the denominator, for example $f(x)=\frac{6 x^4+x^3-7}{2 x^2+8}$, there’s no horizontal asymptote and the limit of the function as $x$ approaches infinity (or negative infinity) does not exist.
If the degree of the denominator is greater than the degree of the numerator, for example $g(x)=\frac{4 x^2-9}{x^3+12}$, the $x$-axis (the line $y=0$ ) is the horizontal asymptote and $\lim {x \rightarrow \infty} g(x)=\lim {x \rightarrow-\infty} g(x)=0$.) If the degrees of the numerator and denominator are equal, take the coefficient of the highest power of $x$ in the numerator and divide it by the coefficient of the highest power of $x$ in the denominator. That quotient gives you the answer to the limit problem and the height of the asymptote. For example, if $h(x)=\frac{4 x^3-10 x+1}{5 x^3+3 x^2-x} \lim {x \rightarrow \infty} h(x)=\lim {x \rightarrow-\infty} h(x)=\frac{4}{5}$, and $h$ has a horizontal asymptote at $y=\frac{4}{5}$.

## MATLAB代写

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