Question

$li{m}_{x\to \frac{\mathrm{\pi }}{2}}\frac{a^{\cot \left(x\right)}-a^{\cos x}}{\cot \left(x\right)-\cos x},a>0=$

• A. ${\log }_e\frac{\mathrm{\pi }}{2}$
• B. ${\log }_{e}2$
• C. ${\log }_{e}a$
• D. $a$

Answer 1

The correct option is C

${\log }_{e}a$

Explanation for Correct Answer:

$limi{t}_{x\to \frac{\mathrm{\pi }}{2}}\frac{a^{\cot \left(x\right)}-a^{\cos x}}{\cot \left(x\right)-\cos x}=$

When we substitute the limit, it becomes the value $\frac{\infty }{\infty }$ does not exist

$$\begin{gathered} limi{t}_{x\to \frac{\mathrm{\pi }}{2}}\frac{a^{\cot \left(x\right)}-a^{\cos x}}{\cot \left(x\right)-\cos x}=limi{t}_{x\to \frac{\mathrm{\pi }}{2}}\frac{a^{\cos x}\left({\displaystyle \frac{a^{\cot \left(x\right)}}{a^{\cos x}}}-1\right)}{\cot \left(x\right)-\cos x}\left[\text{taking}{a}^{\cos x}\text{common in numerator}\right] \\ =limi{t}_{x\to \frac{\mathrm{\pi }}{2}}\frac{a^{\cos x}\left({\displaystyle a^{\cot \left(x\right)-\cos \left(x\right)}-1}\right)}{\cot \left(x\right)-\cos x}\left[\because a^{x-y}=\frac{a^x}{a^y}\right] \\ =limi{t}_{x\to \frac{\mathrm{\pi }}{2}}{a}^{\cos x}.limi{t}_{x\to \frac{\mathrm{\pi }}{2}}\frac{\left({\displaystyle a^{\cot \left(x\right)-\cos \left(x\right)}-1}\right)}{\cot \left(x\right)-\cos x} \end{gathered}$$

$$\begin{gathered} =a^{\cos \frac{\mathrm{\pi }}{2}}.limi{t}_{x\to \frac{\mathrm{\pi }}{2}}\frac{\left({\displaystyle a^{\cot \left(x\right)-\cos \left(x\right)}-1}\right)}{\cot \left(x\right)-\cos x} \\ =a^0.\text{I}\to \left(i\right)\left[\because \cos \frac{\mathrm{\pi }}{2}=0,\text{I}=limi{t}_{x\to \frac{\mathrm{\pi }}{2}}\frac{\left({\displaystyle a^{\cot \left(x\right)-\cos \left(x\right)}-1}\right)}{\cot \left(x\right)-\cos x}\right] \end{gathered}$$

$\text{I}=limi{t}_{x\to \frac{\mathrm{\pi }}{2}}\frac{\left({\displaystyle a^{\cot \left(x\right)-\cos \left(x\right)}-1}\right)}{\cot \left(x\right)-\cos x}$

Let's take $y=\cot \left(x\right)-\cos \left(x\right)$

Limit $x\to \frac{\mathrm{\pi }}{2}$, $\Rightarrow y\to 0\left[\because y=\cot \left(\frac{\mathrm{\pi }}{2}\right)-\cos \left(\frac{\mathrm{\pi }}{2}\right)=0\right]$

$$\begin{gathered} \text{I}=limi{t}_{y\to 0}\frac{\left({\displaystyle a^y-1}\right)}{y} \\ =\ln \left(a\right)\left[\because limi{t}_{y\to 0}\frac{\left({\displaystyle a^y-1}\right)}{y}=\ln \left(a\right)\right] \end{gathered}$$

Substituting the value of $\text{I}$ in equation $\left(i\right)$

$$\begin{gathered} limi{t}_{x\to \frac{\mathrm{\pi }}{2}}\frac{a^{\cot \left(x\right)}-a^{\cos x}}{\cot \left(x\right)-\cos x}=a^0.\text{I} \\ =1.{\log }_{e}a\left[\because a^0=1\right] \end{gathered}$$

Therefore, $limi{t}_{x\to \frac{\mathrm{\pi }}{2}}\frac{a^{\cot \left(x\right)}-a^{\cos x}}{\cot \left(x\right)-\cos x}=$${\log }_{e}a$

Hence, option (B) is the correct answer