Question

For a > 0, let l = $\underset{n\to \frac{\pi }{2}}{lim}\frac{a^{\cot x}-a^{\cos x}}{\cot x-\cos x}$ and m = $\underset{n\to -\infty }{lim}(\sqrt{x^2+ax}-\sqrt{x^2-ax})$ then

• A. 'l' is always greater than 'm' for all values of a> 0
• B. 'l' is always grater than 'm' only when a$\ge$ 1
• C. 'l' is always greater than 'm' for all values of 'a' satisfying a > e${}^{-a}$
• D. e${}^l$ + m = 0

Answer 1

Answer: (C), (D)

e${}^l$ + m = 0
'l' is always greater than 'm' for all values of 'a' satisfying a > e${}^{-a}$

Substitute $x=\frac{\pi }{2}-h$

$l=\underset{h\to 0}{lim}\frac{a^{\tan h}-a^{\sin h}}{\tan h-\sin h}$

$=\underset{h\to 0}{lim}{a}^{\sin h}(\frac{a^{\tan h-\sin h}-1}{\tan h-\sin h})={\log }_{e}a$

$m=\underset{x\to -\infty }{lim}\frac{x^2+ax-x^2+ax}{\sqrt{x^2+ax}+\sqrt{x^2-ax}}$

$=\underset{x\to -\infty }{lim}\frac{2ax}{|x|(\sqrt{1+\frac{a}{x}}+\sqrt{1-\frac{a}{x}})}$

$=\underset{x\to -\infty }{lim}\frac{2ax}{-x\left(2\right)}=-a$