Question

A straight line is said to be an asymptote to the curve y=f(x) if the $\perp$ distance of point P(x, y) on the curve from the line tends to zero when $x\to \infty$ or $y\to \infty$ or both x & y$\to \infty$

(a) Asymptotes parallel to y-axis:

The line $x=k$ is asymptote to the curve y=f(x) If

${lim}_{x\to k^{+}}f\left(x\right)=\infty$ or $-\infty$ & ${lim}_{x\to k^{-}}f\left(x\right)=+\infty$ or $-\infty$

b) Asymptotes parallel to x-axis:

The line $y=k$ is asymptotes to the curve $y=f\left(x\right)$

if ${lim}_{x\to \infty }y=k$ or ${lim}_{x\to -\infty }y=k$

(c) Oblique asymptotes:

Let y=mx+c be an asymptote of the curve y=f(x), then from any point P(x, y) on the curve, the $\perp$ distance is $\frac{|y-mx-c|}{\sqrt{1+m^2}}$ which $\to 0$ as x tends to $+\infty$ or $+\infty$

Thus $y-mx-c\to 0$ as $x\to +\infty$ or $-\infty$

Again $y-mx-c=x(\frac{y}{x}-\frac{c}{x}-m)=0$

As $x\to \infty$ and $y-mx-c\to 0$

So $\underset{x\to \pm \infty }{lim}(\frac{y}{x}-\frac{c}{x}-m)=0$ $\Rightarrow$ $\underset{x\to \pm \infty }{lim}\frac{y}{x}=m$

and ${lim}_{x\to \pm \infty }y-mx-c=0$ $\Rightarrow$ ${lim}_{x\to \pm \infty }y-mx=c$

So we come to the fact that if y=mx+c is an asymptote to the curve y=f(x). Then as $x\to +\infty$ or $-\infty$

${lim}_{x\to \infty }\frac{y}{x}=m={lim}_{x\to \infty }\frac{f\left(x\right)}{x}$ & ${lim}_{x\to \infty }=y-mx={lim}_{x\to \infty }\{f\left(x\right)-mx\}=c$

On the basis of above information answer the following questions. The vertical & horizontal asymptotes to the curve $y=\frac{e^x}{x}$ are respectively

• A. $x=0,y=-1$
• B. $x=0,y=1$
• C. $x=0,y=0$
• D. No asymptote

Answer 1

Answer: (C)

$x=0,y=0$

$\underset{x\to 0^{+}}{lim}y=\underset{x\to 0^{+}}{lim}\frac{e^x}{x}=\left(\underset{x\to 0^{+}}{lim}{e}^x\right)\left(\underset{x\to 0^{+}}{lim}\frac{1}{x}\right)$
=finite number$\times (+\infty )=+\infty$
and $\underset{x\to 0^{-}}{lim}y=\underset{x\to 0^{-}}{lim}\frac{e^x}{x}=\left(\underset{x\to 0^{-}}{lim}{e}^x\right)\left(\underset{x\to 0^{-}}{lim}\frac{1}{x}\right)$
=finite number$\times (-\infty )=-\infty$
$\therefore$ $x=0$ is vertical asymptote
Again $\underset{x\to \infty }{lim}y=\underset{x\to \infty }{lim}\frac{e^x}{x}$($\frac{\infty }{\infty }$ form)
$={lim}_{x\to \infty }{e}^x=\infty$
& $\underset{x\to -\infty }{lim}y=\underset{x\to -\infty }{lim}\frac{e^x}{x}=0$ so $y=0$ is horizontal asymptote