Question

For any real parameter a, let $L\left(a\right)=\underset{x\to 0}{Lt}\frac{(1+x{)}^{\frac{3}{x}}-(1+ax{)}^{\frac{1}{x}}}{x}$. Then which of the following statement(s) is/are true?

• A. L(a) is continuous $\forall aϵR$
• B. L(a) is continuous everywhere except a = 0, 1
• C. $L\left(2\right)=e^2$
• D. $L\left(\frac{1}{2}\right)=\sqrt{e}$
  
  

Answer 1

Answer: (A), (C)

The correct options are
A L(a) is continuous $\forall aϵR$
C $L\left(2\right)=e^2$

Let $y_1=(1+x{)}^{\frac{a}{x}}and{y}_2=(1+ax{)}^{\frac{1}{x}}$
Then $\underset{x\to 0}{Lt}{y}_1=\underset{x\to 0}{Lt}{y}_2=e^a$ .....(i)
Now by L’Hospitals rule
$L\left(a\right)=\underset{x\to 0}{Lt}\frac{y_1^{{}^{\prime }}-y_2^{{}^{\prime }}}{1}$ ....(ii)
Hence we need to take derivatives of the functions and then apply limit
We have, $ln{y}_1=\frac{a}{x}ln(1+x)$. Taking derivative,
$\frac{1}{y_1}{y}_1^{{}^{\prime }}=\frac{a}{x(1+x)}-\frac{a}{x^2}ln(1+x)$ ...(iii)
Similarly, $ln{y}_2=\frac{1}{x}ln(1+ax)$
$\Rightarrow \frac{1}{y^2}{y}_2^{{}^{\prime }}=\frac{a}{x(1+ax)}-\frac{1}{x^2}ln(1+ax)$ ....(iv)
Applying (iii) and (iv) in (ii)
$L\left(a\right)=\underset{x\to 0}{Lt}{y}_1=\left[\frac{a}{x(1+x)-\frac{a}{x^2}ln(1+x)}\right]-y_2[\frac{a}{x(1+ax)}-\frac{1}{x^2}ln(1+ax\left)\right]$
Using (i) as both limits on $y_1$ and $y_2$ are finite and equal, we can factor it out. Hence,
$L\left(a\right)=e^a\underset{x\to 0}{Lt}\frac{a(a-1)x}{(1+x)(1+ax)}-\frac{1}{x^2}(ax-\frac{a{x}^2}{2}+\dots \dots -ax+\frac{a^2{x}^2}{2})$
$=e^a\left[a\right(a-1)-\underset{x\to 0}{Lt}\frac{1}{x^2}(a^2-a\left)\frac{x^2}{2}\right]$
$=e^a\left[a\right(a-1)-\underset{x\to 0}{Lt}-\frac{a(a-1)}{2}]=\frac{a(a-1)}{2}{e}^a$
Hence, L(a) is continuous $\forall aϵR$
$L\left(2\right)=e^2$
$L\frac{1}{8}=-\frac{1}{8}\sqrt{e}$