Limits and Continuity Test 33

Given $$y=f(x)$$ has a unique tangent at the point $$(e^a,0)$$. 
So, as $$x\rightarrow e^{a}$$, $$f(x)\rightarrow 0$$

Now, $$\displaystyle \lim_{x\rightarrow e^{a}}\frac{log_{e}\{1+7f(x)\}-\sin f(x)}{3f(x)}$$

$$\displaystyle \lim _{ x\rightarrow e^{ a } } \frac { log_{ e }(1+7f(x)) }{ 3f(x) } -\frac { \sin f(x) }{ 3f(x) } $$

$$\displaystyle =\frac { 7 }{ 3 } \lim _{ x\rightarrow e^{ a } } \frac { log_{ e }(1+7f(x)) }{ 7f(x) } -\frac { 1 }{ 3 } \lim _{ x\rightarrow e^{ a } } \frac { \sin f(x) }{ f(x) } $$

$$=\dfrac{7-1}{3}=2$$

Given: $$\displaystyle f\left( x \right) =\begin{cases} \begin{matrix} \dfrac { \cos ^{ 2 }{ x } -\sin ^{ 2 }{ x-1 }  }{ \sqrt { { x }^{ 2 }+4 } -2 } , & \left( x\neq 0 \right)  \end{matrix} \\ \begin{matrix} a, & \left( x\neq 0 \right)  \end{matrix} \end{cases}$$ is continuous at $$x=0$$, then 

$$f(x)=\displaystyle \frac { sin3x+Asin2x+Bsinx }{ x^{ 5 } } ;x\neq 0$$
$$\displaystyle \lim _{ x\rightarrow 0 }{ f(x)= } \lim _{ x\rightarrow 0 }{\displaystyle \frac { sin3x+Asin2x+Bsinx }{ x^{ 5 } }  } $$
It is of the  form $$\displaystyle \frac{0}{0}$$, so applying L-Hospital's rule
$$=\displaystyle \lim _{ x\rightarrow 0 }{ \displaystyle\frac { 3\cos { 3x } +2A\cos { 2x } +B\cos { x }  }{ 5x^{ 4 } }  } $$
As $$x\to 0 , D^{r} \to 0 \Rightarrow N^{r} \to 0$$
$$\displaystyle \lim _{ x\rightarrow 0 }{ 3\cos { 3x } +2A\cos { 2x } +B\cos { x } =0 } $$
$$\Rightarrow 3+2A+B=0$$            .....(i)
Again $$\displaystyle \lim _{ x\rightarrow 0 }{\displaystyle \frac { 3\cos { 3x } +2A\cos { 2x } +B\cos { x }  }{ 5x^{ 4 } }  } $$ is of the form $$\displaystyle \frac {0}{0}$$
$$=\displaystyle \lim _{ x\rightarrow 0 }{ \displaystyle\frac { -9\sin { 3x } -4A\sin { 2x } -B\sin { x }  }{ 20x^{ 3 } }  } $$
Again of the form $$\displaystyle \frac {0}{0}$$
$$=\displaystyle \lim _{ x\rightarrow 0 }{\displaystyle \frac { -27\cos { 3x } -8A\cos { 2x } -B\cos { x }  }{ 60x^{ 2 } }  } $$
As $$x\to 0 , D^{r} \to 0 \Rightarrow N^{r} \to 0$$
$$\displaystyle \lim _{ x\rightarrow 0 }{ -27\cos { 3x } -8A\cos { 2x } -B\cos { x }  } =0$$
$$\Rightarrow 27+8A+B=0$$            .....(ii)
Solving (i) and (ii), we get
$$A=-4, B=5$$
Thus, $$A+B=1$$

We have,
$$\displaystyle \lim_{n \rightarrow \infty} x^{2n} = \left\{\begin{matrix}0,& if  |x| < 1\\ \infty, & if  |x| > 1 \\ 1, & if |x| = 1\end{matrix}\right.$$
Thus, we have the following cases:
CASE I When $$- 1 < x < 1$$
In this case, we have $$\displaystyle \lim_{n \rightarrow \infty} x^{2n} = 0$$
$$\therefore \phi (x) =\displaystyle  \lim_{n \rightarrow \infty} \frac{x^{2n} f(x) + g(x)}{1 + x^{2n}} = g(x)$$
CASE II When $$|x| > 1$$
In this case, we have $$\displaystyle \lim_{n \rightarrow \infty} \displaystyle \frac{1}{x^{2n}} = 0$$
$$\therefore \phi (x) = \displaystyle \lim_{n \rightarrow \infty} \frac{x^{2n} f(x) + g(x)}{1 + x^{2n}}$$
$$\Rightarrow \phi (x) =\displaystyle  \lim_{n \rightarrow \infty} \frac{f(x) + \displaystyle \frac{g(x)}{x^{2n}}}{1 + \displaystyle \frac{1}{x^{2n}}} = \displaystyle \frac{f(x) + 0}{1 + 0} = f(x)$$
CASE III  When $$|x| = 1$$
In this case, we have $$x^{2n} = 1\Rightarrow \displaystyle \lim_{n \rightarrow \infty } x^{2n}=1$$.
$$\therefore \phi (x) = \displaystyle \frac{f(x) + g(x)}{2}$$

$$\displaystyle \lim_{h \rightarrow 0} \frac{x- e^x + 1 - (1 - cos  2x)}{x^2}$$

Given f(x) is continuous at $$x=0 $$
$$\Rightarrow LHL=RHL=f(0)$$

Now, $$f(0) = 0 + 0 + \lambda  ln  4= \lambda  ln  4$$                 ....(1)
$$R.H.L. = \displaystyle \lim_{x  \rightarrow 0^+} f(x) = \lim_{h \rightarrow 0} f(0+h)$$

I) $$\displaystyle \lim_{x \rightarrow 0} [x] \left \{ \frac{e^{1/x} - 1}{e^{1/x} + 1} \right \}$$

$$RHL=\displaystyle \lim_{x \rightarrow 0^+} [x] \left ( \frac{e^{1/x} - 1}{e^{1/x}+1} \right) $$

$$=\lim _{ h\rightarrow 0 } [h]\left(\displaystyle \frac { e^{ 1/h }-1 }{ e^{ 1/h }+1 }  \right) $$

$$= \lim_{h  \rightarrow 0} [h] \left ( \displaystyle\frac{1 - e^{-1/h}}{1 + e^{-1/h}} \right) $$
$$= 0 \times 1 = 0$$

$$LHL=\displaystyle \lim_{x \rightarrow 0^-} [x] \left ( \frac{e^{1/x}- 1}{e^{1/x} + 1} \right )$$

$$= \lim_{h \rightarrow 0} [-h] \left ( \displaystyle\frac{e^{-1/h} - 1}{e^{-1/h} + 1} \right) $$
$$= -1 \times (-1) = 1$$
Here, $$LHL \ne RHL$$
Thus, given limit does not exist.

II) $$\displaystyle \lim_{x \rightarrow 0} \left ( \frac{e^{1/x} - 1}{e^{1/x} + 1} \right )$$ 

$$RHL=\displaystyle \lim_{x \rightarrow 0^+} \left ( \frac{e^{1/x} - 1}{e^{1/x}+1} \right) $$

$$=\lim _{ h\rightarrow 0 }\left(\displaystyle \frac { e^{ 1/h }-1 }{ e^{ 1/h }+1 }  \right) $$

$$= \lim_{h  \rightarrow 0} \left ( \displaystyle\frac{1 - e^{-1/h}}{1 + e^{-1/h}} \right) $$
$$RHL= 1 $$

$$LHL=\displaystyle \lim_{x \rightarrow 0^-} \left ( \frac{e^{1/x}- 1}{e^{1/x} + 1} \right )$$

$$= \lim_{h \rightarrow 0}  \left ( \displaystyle\frac{e^{-1/h} - 1}{e^{-1/h} + 1} \right) $$
$$LHL= -1$$

So, $$\displaystyle \lim_{x \rightarrow 0} \left ( \frac{e^{1/x} - 1}{e^{1/x} + 1} \right)$$ does not exist, but this cannot be taken as only reason for non-existence of $$\displaystyle \lim_{x \rightarrow 0} [x] \left ( \frac{e^{1/x} - 1}{e^{1/x} + 1} \right )$$.