Question

Evaluate: The Limit of x to infinity 2^x + x^3/x^2 + 3^x

My choices are 0, 1, 3/2, 2/3, and The limit does not exist.

However I'm pretty sure that the answer is 2/3

Answer 1

and btw i think you're right about your answer :)

Answer 2

Okay, Thank you! :) I just wanted to make sure

Answer 3

No that is not the right answer

Answer 4

Okay, How would I calculate the correct answer?

Answer 5

Let me convince you that the limit is 0

Answer 6

The only way I could see the limit being 0 is if we look at the bases

Answer 7

\[
2^x + x^3\approx 2^x
\]
when x is near \( \infty\)

Answer 8

\[
x^2 + 3^x\approx 3^x

\]
when x is near \( \infty\)

Answer 9

The ratio is like
\[
\frac{2^x}{3^x}= \left ( \frac 2 3 \right)^x \to 0\]
when x goes to \(\infty \)

Answer 10

Exponential growth is the deciding part near \( \infty \)

Answer 11

Ohhhh Okay, and in my lecture they looked at bases when it came to exponential functions. So because 3 is the larger base and is in the denominator, it automatically goes to 0, I see now. Thank you for explaining that!!

Answer 12

A more formal proof goes as
\[
\frac{x^3+2^x}{x^2+3^x}=\frac{x^3+2^x}{\frac{3^x
\left(x^2+3^x\right)}{3^x}}=\frac{\frac{x^3}{3^x}+\left(\frac{2}{3}\right)^
x}{\frac{x^2}{3^x}+1}\]
and let x goes to \(\infty\)

Answer 13

I divided up and down by \[ 3^x

\]

Answer 14

YW