# Is this workaround used by my book to find the value of $1/0$ legitimate? [duplicate]

I was doing a physics problem and found that I reached an interesting equation:-

To cross the river directly from A to B (making a right angle with the velocity of the stream), how must the fisherman direct his/her boat's motion? Find the angle ($$\alpha$$) between the velocity vector of the stream and the resultant velocity vector of the boat. The magnitude of the velocity of the boat is $$v=2.778ms^{-1}$$ and that of the stream is $$u=1.3889ms^{-1}$$.

Solution:

We know from the Parallelogram law of addition of vectors:

$$\tan\theta=\frac{Qsin\alpha}{P+Qcos\alpha}$$

Similarly,

$$\tan(90^{\circ})=\frac{v\sin\alpha}{u+v\cos\alpha}$$

$$\implies\frac{\sin(90^{\circ})}{\cos(90^{\circ})}=\frac{v\sin\alpha}{u+v\cos\alpha}$$

$$\implies \frac{1}{0}=\frac{v\sin\alpha}{u+v\cos\alpha}$$

$$\implies{u+v\cos\alpha}=0$$

Now, isn't $$\;\dfrac10\;$$ undefined? How is cross-multiplication valid here? I'm very confused.

• it is not valid however as any fraction with finite numerator and denominator approaches infinity, the denominator must approach $0$ Jun 17, 2021 at 14:44
• That representation of $\tan 90$ is not valid (mathematically, I won't speak for physicists) and the cross-multiplication is also invalid. What you could rigorously say is that since $\tan \theta$ grows without bound as $\theta \rightarrow 90^\circ$ and $\sin \alpha$ is bounded, either $v+v \cos \alpha$ must shrink to $0$ or $u$ grows without bound. Assuming you have constraints on $u$ you're then (mostly) in the position of your last line Jun 17, 2021 at 14:44
• You're not confused; you're correct. Cross multiplication is not legitimate here. And writing $1/0$ is meaningless. What we do know is that as the argument of the tangent function appoaches $90$ degrees from the left, the term on the right-hand side tends to $\infty$. The only way for this to happen is that the denominator tends to $0$ since the speed $u$ is bounded by the speed of light. Jun 17, 2021 at 14:45
• One line answer: $\tan 90$ is not $\frac{1}{0}$ I am confused whether to write this as a answer as it is or rather leave it a comment
– user876009
Jun 17, 2021 at 14:47
• @postmortes ya I also thought same so rather left it as a comment
– user876009
Jun 17, 2021 at 14:48