Search any question & find its solution
Question:
Answered & Verified by Expert
Let $a>0$ be a real number. Then the limit $\lim _{x \rightarrow 2} \frac{a^{x}+a^{3-x}-\left(a^{2}+a\right)}{a^{3-x}-a^{x / 2}}$
is
Options:
is
Solution:
2980 Upvotes
Verified Answer
The correct answer is:
$\frac{2}{3}(1-a)$
$\lim _{x \rightarrow 2} \frac{a^{x}+a^{3-x}-\left(a^{2}+a\right)}{a^{3-x}-a^{x / 2}}$
$\lim _{x \rightarrow 2} \frac{\left(a^{x}-a\right)\left(a^{x / 2}-a\right)\left(a^{x / 2}+a\right)}{\left(a-a^{x / 2}\right)\left(a^{2}+a^{x}+a^{x / 2} \cdot a\right)}$
$\lim _{x \rightarrow 2}-\frac{\left(a^{x}-a\right)\left(a^{x / 2}+a\right)}{a^{2}+a^{x}+a^{x / 2} \cdot a}$
$=\frac{2}{3}(1-a)$
$\lim _{x \rightarrow 2} \frac{\left(a^{x}-a\right)\left(a^{x / 2}-a\right)\left(a^{x / 2}+a\right)}{\left(a-a^{x / 2}\right)\left(a^{2}+a^{x}+a^{x / 2} \cdot a\right)}$
$\lim _{x \rightarrow 2}-\frac{\left(a^{x}-a\right)\left(a^{x / 2}+a\right)}{a^{2}+a^{x}+a^{x / 2} \cdot a}$
$=\frac{2}{3}(1-a)$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.