Question

$\displaystyle \lim_{x \to 2}$ $\left(\frac{e^{x}-e^{2}}{x-2}\right)=$

• A. $0$
• B. $1$
• C. $e^2$
• D. $e$

Answer 1

Answer: (C)

Put $x - 2 = y$, so that when $x \to 2$ then $y \to 0$.
$\therefore \displaystyle \lim_{x \to 2}$ $\left(\frac{e^{x}-e^{2}}{x-2}\right)$
$=\displaystyle \lim_{y \to 0}$$\left(\frac{e^{y+2}-e^{2}}{y}\right)$
$=\displaystyle \lim_{y \to 0}$$\left\{e^{2}\cdot\left(\frac{e^{y}-1}{y}\right)\right\}$
$=e^{2}\cdot\displaystyle \lim_{y \to 0}$$\left(\frac{e^{y}-1}{y}\right)$
$=e^{2}\times1=e^{2}$