Computing limits - logs vs powers

In summary, the limit of (log(x^2+e^(2x))/(x+3)) as x approaches infinity is 2. This is because e^2x dominates the numerator and can be simplified to just 2x, and x^2+e^2x can be bounded by 2e^2x. Therefore, the limit can be sandwiched between 2 and 2e^2x, which both tend to 2 as x approaches infinity.
  • #1
wombat4000
36
0

Homework Statement



[tex]\frac{log (x^{2} + e^{2x})}{x + 3}[/tex] find the limit as x_> infinty

Homework Equations



powers beat logs

The Attempt at a Solution



going by the powers beat logs idea - simply, the limit as n-> infinity is 0.

is this correct? can you simply say that powers beat logs always?
 
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  • #2
No it's not that simple. Intuitively e^2x is going to dominate the bracket in the numerator and log(e^2x) can be easily simplified. Can you sandwich x^2+e^2x between multiples of e^2x (at least for large x)? Try the sandwich theorem.
 
  • #3
ok - it will go to 2 then?

[tex]\frac{log (e^{2x})}{x + 3} \leq \frac{log (x^{2} + e^{2x})}{x + 3}
\leq ? [/tex]
[tex]\frac{log (e^{2x})}{x + 3} = \frac{2x}{x + 3} [/tex] which tends to 2 as x tends to infinity is this correct - what should i use for thew upper bound?
 
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  • #4
Fine so far. How would you compare x^2 and e^2x? (at least for large x).
 
  • #5
e^2x is considerably larger than x^2 ?

if x^2 divereges - so must e^2x?

cant say I am 100% sure what your ginting towards.
 
  • #6
Sorry if I'm being too vague.

x^2<e^2x so x^2+e^2x<2e^2x which gives you another useful bound :)
 
  • #7
brilliant! - thanks

http://img223.imageshack.us/img223/2827/167uqnbqe2.gif
 
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FAQ: Computing limits - logs vs powers

What is the difference between computing limits using logs and powers?

The main difference is that when computing limits using logs, the logarithmic function is used, whereas when computing limits using powers, the power function is used. This means that the approach and method of calculation may differ. Additionally, the values of the limits may also differ depending on which method is used.

When should I use logs to compute limits?

Logs are useful for computing limits when the function being evaluated has a complex or indeterminate form, such as 0/0 or infinity/infinity. In these cases, taking the logarithm of the function may simplify the expression and make it easier to evaluate the limit.

Can I always use logs to compute limits?

No, logs may not always be the best approach for computing limits. In some cases, using powers may be more efficient and yield a more accurate result. It is important to understand the properties and limitations of both logs and powers when deciding which method to use for computing a limit.

How do I know if I should use logs or powers to compute a limit?

Choosing between logs and powers for computing a limit depends on the specific function being evaluated and its properties. Generally, if the function has a logarithmic form, then using logs may be a good approach. Otherwise, using powers may be more suitable.

Are there any common mistakes to avoid when using logs or powers to compute limits?

One common mistake is forgetting to apply the properties of logs or powers when evaluating a limit. It is important to have a thorough understanding of these properties and how they affect the limit calculation. Another mistake is relying solely on one method (logs or powers) and not considering alternative approaches, which may be more efficient or accurate.

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