Limits and Continuity Test 14

 $$\displaystyle \lim_{x\to0}{\displaystyle \frac{x^n - \sin^nx}{x - \sin^nx}}$$

$$\mbox{For n = 0,we have }\displaystyle\lim_{x\to0}\displaystyle\frac{1-1}{x-1}=0$$

$$\mbox{For n = 1,}\displaystyle\lim_{x\to0}\displaystyle\frac{x-\sin x}{x- \sin x}=1\\$$

$$\mbox{For n = 2,}\displaystyle\lim_{x\to0}\displaystyle\frac{x^2-\sin^2x}{x-\sin^2x}=\displaystyle\lim_{x\to0}\displaystyle\frac{1-\displaystyle\frac{\sin^2x}{x^2}}{\displaystyle\frac{1}{x}-\displaystyle\frac{\sin^2x}{x^2}}.\\$$
$$\mbox{This does not exist.}\\ $$

$$\displaystyle { For\ \  n=3, }\lim _{ x\to 0 } \dfrac { x^{ 3 }-\sin ^{ 3 } x }{ x-\sin ^{ 3 } x } =\lim _{ x\to 0 } \dfrac { 1-\dfrac { \sin ^{ 3 } x }{ x^{ 3 } }  }{ \dfrac { 1 }{ x^{ 2 } } -\dfrac { \sin ^{ 3 } x }{ x^{ 3 } }  } $$
For  $$n = 3$$, also given limit does not exist.
Hence, $$n = 1$$

$$\displaystyle \lim_{n \rightarrow \infty} \dfrac{-3n + (-1)^n}{4n - (-1)^n}$$

$$\displaystyle =\lim_{n \rightarrow \infty} \dfrac{-3 + \dfrac{(-1)^n}{n}}{4 + \dfrac{(-1)^n}{n}} = \dfrac{-3}{4}$$

$$\displaystyle\lim_{x\to1}\frac{1-x^2}{\sin2\pi
x}= -\displaystyle \lim_{x\to1}\frac{2\pi(1-x)(1+x)}{-2\pi\sin(2\pi-2\pi x)} $$

$$=\displaystyle \lim_{x \to 1}\left( \frac{(2\pi-2\pi x}{\sin(2\pi-2\pi x)}\right).\frac{1+x}{-2\pi}=-\frac{1}{\pi}$$

$$\displaystyle\lim_{x\to\infty}\left(\frac{x^3+1}{x^2+1}-(ax+b)\right)= 2\\$$

$$or\displaystyle\lim_{x\to\infty}\frac{x^3(1-a)-bx^2-ax+(1-b)}{x^2+1} = 2\\$$

As we have finite limit so degree of $$x$$ in the numerator  must be less than equal to the degree of $$x$$ in the denominator.

So the term containing $$x^3$$ must have coefficient $$0$$ i.e $$(1-a) = 0$$

Hence $$a =1$$.Divide Numerator and denominator by $$x^3$$ to get the value of $$b$$

$$ -b = 2$$

$$\mbox{a = 1,b = -2}\\$$

Hence, option 'C' is correct.

$$\displaystyle \lim_{x \rightarrow 1^+} f(g(x)) = f(g(1^+)) = f((2^+)) = 2^2 + 2 = 6$$
and $$\displaystyle \lim_{x - 1^-} f(g(x)) = f(g(1^-)) = f(3 - 1^-) =f(2^+) = 2^2 + 2 = 6$$
Hence, $$\displaystyle \lim_{x \rightarrow 1} f(g(x)) = 6$$.

$$\displaystyle f(x) = \frac{3x^2 + ax + a + 1}{(x + 2)(x-1)}$$
As $$x \rightarrow -2, D^r \rightarrow 0$$. Hence, as $$x  \rightarrow -2, N^r \rightarrow 0$$. Therefore,
$$\therefore 12 - 2a + a + 1 = 0$$ or $$a = 13$$
Hence, option 'A' is correct.

$$\displaystyle\lim_{x\to\infty}\displaystyle\frac{\sin3x}{x^3}+\displaystyle\frac{a}{x^2}+b

= \lim_{x\to0}\displaystyle\frac{\sin3x+ax+bx^3}{x^3}$$

$$\displaystyle\lim_{x\to\infty}\left(\displaystyle  \frac{x^2+2x-1}{2x^2-3x-2}\right)^{\displaystyle  \frac{2x+1}{2x-1}}\\$$

$$=\displaystyle\lim_{x\to\infty}\left(\displaystyle \frac{1+\displaystyle  \frac{2}{x}-\displaystyle  \frac{1}{x^2}}{2-\displaystyle  \frac{3}{x}-\displaystyle  \frac{2}{x^3}}\right)^{\displaystyle \frac{2+{{1}/{x}}}{2-{{1}/{x}}}}\\$$

$$=1/2$$

We have,
$$\displaystyle \lim_{x \rightarrow 1} \frac{x^8 -2x + 1}{x^4 - 2x + 1}$$