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

[tandoorichicken]

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 $$\lim_{t\rightarrow k} \frac{F(t)}{G(t)+7}$$?

 

[Jupiter]

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).

 

[M98Ranger]

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!