Question

i. find lim𝑥→2 𝑥^ 2−3𝑥 +2 / 𝑥 ^2−4

ii. Where are is the function 𝑓(𝑥) = 1 2𝑥 ^2−6𝑥+4 continuous?

iii. Given that given 𝑌 = tan−1 𝑥 show that 𝑑𝑦 /𝑑𝑥 = 1 1+𝑥 ^2 

iv. Prove using the first principle that the derivative of sin 𝑥 is cos 𝑥 and that the derivative of 𝑐𝑜𝑠𝑥 is – 𝑠𝑖𝑛x

v. Differentiate sin 𝑦 − 𝑥 2𝑦 3 − 𝑐𝑜𝑠𝑥 = 3y

vi. lim𝑥→1 𝑥^2+𝑥 3−5𝑥+3/ 𝑥^3+2𝑥^2+7𝑥+4

vii. lim𝑥→∞ 2𝑥/ 3𝑥^6+𝑥+4

Answer 1

i.

$$\lim\limits_{x\to 2}\dfrac{x^2-3x+2}{x^2-4}=\lim\limits_{x\to 2}\dfrac{(x-1)(x-2)}{(x-2)(x+2)}$$

$$=\lim\limits_{x\to 2}\dfrac{x-1}{x+2}=\dfrac{2-1}{2+2}=\dfrac{1}{4}$$

ii. The function $f(x)=12x^2-6x+4$ is continuous for all $x\in \R$ as polynomial.

iii. Let $y=\tan^{-1}(x).$ Then $\tan y=x, -\dfrac{\pi}{2}<y<\dfrac{\pi}{2}$

$$(\tan^{-1}(x))'=\dfrac{1}{(\tan y)'}=\dfrac{1}{\dfrac{1}{\cos^2y}}$$

$$=\dfrac{1}{1+\tan^2y}=\dfrac{1}{1+x^2}$$

iv.

$$(\sin x)'=\lim\limits_{h\to 0}\dfrac{\sin(x+h)-\sin x}{h}$$

$$=\lim\limits_{h\to 0}\dfrac{\sin x\cos h+\cos x \sin h-\sin x}{h}$$

$$=\lim\limits_{h\to 0}\dfrac{\sin x(\cos h-1)+\cos x \sin h}{h}$$

$$=\sin x\lim\limits_{h\to 0}\dfrac{\cos h-1}{h}+\cos x\lim\limits_{h\to 0}\dfrac{\sin h}{h}$$

$$=\sin x(0)+\cos x(1)=\cos x$$

$$(\cos x)'=\lim\limits_{h\to 0}\dfrac{\cos(x+h)-\cos x}{h}$$

$$=\lim\limits_{h\to 0}\dfrac{\cos x\cos h-\sin x \sin h-\cos x}{h}$$

$$=\lim\limits_{h\to 0}\dfrac{\cos x(\cos h-1)-\sin x \sin h}{h}$$

$$=\cos x\lim\limits_{h\to 0}\dfrac{\cos h-1}{h}-\sin x\lim\limits_{h\to 0}\dfrac{\sin h}{h}$$

$$=\cos x(0)-\sin x(1)=-\sin x$$

v.

$$\dfrac{d}{dx}(\sin y-x^2y^3-\cos x)=\dfrac{d}{dx}(3y)$$

Use the Chain Rule

$$\cos y \dfrac{dy}{dx}-2xy^3-3x^2 y^2 \dfrac{dy}{dx}+\sin x=3\dfrac{dy}{dx}$$

Solve for $\dfrac{dy}{dx}$

$$\dfrac{dy}{dx}=\dfrac{-2xy^3+\sin x}{3+3x^2y^2-\cos y}$$

vi.

$$\lim\limits_{x\to 1}\dfrac{x^2+x^3-5x+3}{x^3+2x^2+7x+4}=\dfrac{1^2+1^3-5(1)+3}{1^3+2(1)^2+7(1)+4}$$

$$=0$$

vii.

$$\lim\limits_{x\to \infin}\dfrac{2x}{3x^6+x+4}=\lim\limits_{x\to \infin}\dfrac{\dfrac{2x}{x^6}}{\dfrac{3x^6}{x^6}+\dfrac{x}{x^6}+\dfrac{4}{x^6}}$$

$$=\lim\limits_{x\to \infin}\dfrac{\dfrac{2}{x^5}}{3+\dfrac{1}{x^5}+\dfrac{4}{x^6}}=\dfrac{0}{3+0+0}=0$$