Solving Limits with Functions: \lim_{t\rightarrow k}

  • Thread starter Thread starter tandoorichicken
  • Start date Start date
  • Tags Tags
    Functions Limits
AI Thread Summary
The discussion focuses on solving limits involving functions F(t) and G(t) as t approaches k. It establishes that if the limit of F(t) is 7 and G(t) is 0, then the limit of their product, F(t)G(t), is 0. For the quotient limit, since G(t) approaches 0, the limit of G(t) + 7 approaches 7, allowing the limit of the quotient to be calculated as 1. The discussion emphasizes the importance of limit laws and encourages practice for mastery. Understanding these concepts is crucial for solving similar limit problems effectively.
tandoorichicken
Messages
245
Reaction score
0
I forgot how to do these kind of problem:

If \lim_{t\rightarrow k} F(t) = 7 and \lim_{t\rightarrow k} G(t) = 0, then what is \lim_{t\rightarrow k} F(t)G(t)?

Also:

What is \lim_{t\rightarrow k} \frac{F(t)}{G(t)+7}?
 
Physics news on Phys.org
Since both the limits exist, the limit of the product is just the product of the limits. Also, if you have a quotient and the limit on the top exists and the limit on the bottom exists and is nonzero, then the limit of the quotient is just the quotient of the limits.
If \lim_{t\rightarrow k} G(t)=0, then \lim_{t\rightarrow k} G(t)+7=7, since the limit of a sum is just the sum of the limits (if both limits exist).
 



Don't worry, solving limits with functions can be a bit tricky at first but with practice, you'll get the hang of it again. To solve the first problem, we can use the limit laws which state that the limit of a product is equal to the product of the limits. So, in this case, we can say that \lim_{t\rightarrow k} F(t)G(t) = \lim_{t\rightarrow k} F(t) \cdot \lim_{t\rightarrow k} G(t) = 7 \cdot 0 = 0. Therefore, the limit of the product is 0.

For the second problem, we can use another limit law which states that the limit of a quotient is equal to the quotient of the limits, as long as the limit of the denominator is not 0. In this case, the limit of the denominator is 0, so we cannot use this law. Instead, we can try to factor out a common factor from the numerator and denominator to eliminate the 0 in the denominator.

\lim_{t\rightarrow k} \frac{F(t)}{G(t)+7} = \lim_{t\rightarrow k} \frac{F(t)}{G(t)} \cdot \frac{1}{1+\frac{7}{G(t)}}

Since we know that \lim_{t\rightarrow k} F(t) = 7 and \lim_{t\rightarrow k} G(t) = 0, we can rewrite this as:

= \frac{7}{0+7} = 1

Therefore, the limit is equal to 1. Keep practicing and don't be afraid to ask for help if you get stuck. Good luck!
 
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top