Answers to Calculus Practical 4

Question 1 §

Question 1a §

$$\underset{x\to 0}{\lim }\frac{x^2+2x+1}{x+1}=1$$

Question 1b §

$$\underset{x\to 2}{\lim }\frac{x^2-5x+6}{x^2-4}:\text{undefined at x=2, so use L’Hoˆpital’s Rule}$$

$$⟹\underset{x\to 2}{\lim }\frac{2x-5}{2x}=-\frac{1}{4}$$

Question 2 §

Question 2a §

$$\underset{x\to 0}{\lim }(x\cos x)=0$$

Question 2b §

$$\underset{x\to 0}{\lim }\frac{\sin x}{e^{2x}-e^{-2x}}:\text{undefined at x=0, so use L’Hoˆpital’s Rule}$$

$$⟹\underset{x\to 0}{\lim }\frac{\cos x}{2e^{2x}+2e^{-2x}}=\frac{1}{4}$$

Question 2c §

$$\underset{x\to 0}{\lim }\frac{1-\cos 2x}{x{e}^x-x}:\text{undefined at x=0, so use L’Hoˆpital’s Rule}$$

$$⟹\underset{x\to 0}{\lim }\frac{2\sin 2x}{e^x+x{e}^x-1}:\text{undefined at x=0, so use L’Hoˆpital’s Rule}$$

$$⟹\underset{x\to 0}{\lim }\frac{4\cos 2x}{2e^x+x{e}^x}=2$$

Question 2d §

$$\underset{y\to 0}{\lim }\frac{\tan y}{\tan ay},\text{ where }a\text{ is a non-zero constant}:\text{undefined at y=0, so use L’Hoˆpital’s Rule}$$

$$\underset{y\to 0}{\lim }\frac{{\sec }^{2}y}{a{\sec }^{2}ay}=\frac{1}{a}$$

Question 2e §

$$\underset{t\to 5}{\lim }\frac{2-\sqrt{t-1}}{t^2-25}\text{undefined at t=5, so use L’Hoˆpital’s Rule}$$

$$\underset{t\to 5}{\lim }\frac{-\frac{1}{2}(t-1{)}^{-\frac{1}{2}}}{2t}=-\frac{1}{40}$$

Question 3 §

Question 3a §

$$\underset{x\to \infty }{\lim }\frac{x^4-2x^3+3x^2+5}{4x^4-1}=\frac{1}{4}\text{, by evaluating the ratio between the highest order }x$$

Question 3b §

$$\underset{v\to \infty }{\lim }\frac{5v^9-4v^5+3v-21}{-2v^9+v^8-4v^2+3v}=-\frac{5}{2}\text{, by evaluating the ratio between the highest order }v$$

Question 3c §

$$\underset{u\to \infty }{\lim }\frac{u^3-3u^2+5}{u^4+5u^3+1}=0\text{, by evaluating the ratio between the highest order }u$$

Question 4 §

Question 4a §

$$f(x)=x\sin \left(\frac{1}{x}\right):\text{continuous for the interval }(-\infty ,0)\cup (0,\infty )\text{, as shown below.}$$

As $x$ approaches $0$, the function oscillates infinitely, making it hard to predict its behaviour at $x=0$, thus it’s an oscillating discontinuity.

However, the $x$ multiplying the sine function ensures that it is equal to 0 ($0\cdot a=0$, even if $a$ is undefined). So it is continuous for all real values of $x$.

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f(x)=x*sin(1/x)

Question 4b §

$$F(x)=\frac{x^2-3x+2}{x-2}:\text{continuous for the interval }(-\infty ,2)\cup (2,\infty )\text{, as shown below.}$$

$$F(2)=\frac{2^2-3\cdot 2+2}{2-2}=\text{undefined}$$

However, if we simplify the function by factorising $x^2-3x+2⟹(x-2)(x-1)$, then function simplifies to just $x-1$. Thus, it’s a removable discontinuity.

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f(x)=(x^2-(3*x)+2)/(x-2)

Question 4c §

$g(x)=\tan x$ The tangent function is continuous everywhere except where its denominator, $\cos x$, is zero. This occurs at odd multiples of $\frac{\pi }{2}$. Therefore, $g(x)$ is continuous for all $x$ except at $x=(2n+1)\frac{\pi }{2}$ where $n$ is any integer. I wasn’t sure how to write this in interval notation.

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g(x)=tan(x)

Question 4d §

$$G(x)=\cos (x^3+2x^2+3):\text{continuous for the interval }(-\infty ,\infty )\text{, as shown below.}$$

The cosine function is continuous everywhere. Since the polynomial inside the cosine function is also continuous everywhere, $G(x)$ is continuous for all real values of $x$.

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G(x)=cos(x^3+2x^2+3)

Question 4e §

$$h(x)=x^{500}-2x^6+80$$

This is a polynomial, and polynomials are continuous everywhere. Therefore, $h(x)$ is continuous for all real values of $x$.

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h(x)=x^500-2x^6+80

Question 4f §

$$H(x)=\frac{x^2+2x+3}{x^2-1}:\text{continuous for the interval }(-\infty ,-1)\cup (-1,1)\cup (1,\infty )\text{, as shown below.}$$

This is a rational function. Rational functions are continuous wherever their denominator is not zero. In this case, the denominator $x^2-1$ is zero when $x=1$ and $x=-1$. Therefore, $H(x)$ is continuous for all $x$ except at $x=1$ and $x=-1$.

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H(x)=(x^2+2x+3)/(x^2-1)

Question 5 §

$$g(x)=\underset{x\to \infty }{\lim }\frac{\sin x}{x}$$

The Squeeze Theorem dictates that: If $f(x)\le g(x)\le h(x)$ for all $x$ in some open interval containing $a$ (except possibly at $a$), and: ${\lim }_{x\to a}f(x)={\lim }_{x\to a}h(x)=L$, then: ${\lim }_{x\to a}g(x)=L$.

Thus, if we find $f(x)$ and $h(x)$ as defined, we can find the limit of $g(x)$.

We know that $\sin x$ is defined as $-1\le \sin x\le 1$, so if we divide through by $x$ then:

$$-\frac{1}{x}\le \frac{\sin x}{x}\le \frac{1}{x}⟹f(x)\le g(x)\le h(x)$$

Thus the limit of $g(x)$ is the same as the limit of $h(x)$:

$$\underset{x\to \infty }{\lim }h(x)=\underset{x\to \infty }{\lim }\frac{1}{x}=0$$

$$⟹\underset{x\to \infty }{\lim }\frac{\sin x}{x}=0\text{ by Squeeze Theorem}■$$

Question 6 §

Question 6a §

$$\underset{x\to 0}{\lim }\left(\frac{1}{\sin x}-\frac{1}{x}\right)=\underset{x\to 0}{\lim }\left(\frac{1}{\sin x}\right)-\underset{x\to 0}{\lim }\left(\frac{1}{x}\right)=\infty -\infty =0$$

I know this is incorrect notation, but I don’t know the correct version.

Question 6b §

$$\underset{\theta \to \frac{\pi }{2}}{\lim }(\sec \theta -\tan \theta )=\underset{\theta \to \frac{\pi }{2}}{\lim }(\sec \theta )-\underset{\theta \to \frac{\pi }{2}}{\lim }(\tan \theta )=\frac{1}{{\lim }_{\theta \to \frac{\pi }{2}}(\cos \theta )}-\frac{{\lim }_{\theta \to \frac{\pi }{2}}(\sin \theta )}{{\lim }_{\theta \to \frac{\pi }{2}}(\cos \theta )}$$

$$=\frac{1}{{\lim }_{\theta \to \frac{\pi }{2}}(\cos \theta )}-\frac{1}{{\lim }_{\theta \to \frac{\pi }{2}}(\cos \theta )}=0$$

Question 6c §

$$\underset{h\to 0}{\lim }\frac{\sqrt[5]{32+h}-2}{h}=\underset{h\to 0}{\lim }\frac{\sqrt[5]{32+h}-\sqrt[5]{32}}{h}:f^{\prime }(x)=\underset{h\to 0}{\lim }\frac{f(x+h)-f(x)}{h}$$

Thus, evaluation of the limit is equal to differentiating the function $f(x)=\sqrt[5]{x}=x^{\frac{1}{5}}$ at the point $x=32$:

$$f^{\prime }(x)=\frac{1}{5}{x}^{-\frac{4}{5}}⟹f^{\prime }(32)=\frac{1}{5}\cdot \frac{1}{2^4}=\frac{1}{80}$$