MHB Calculating Limits without Cheating or L'Hopital's Rule

[Yankel]

Hello all,

I am trying to calculate the following limits, without cheating and using a calculator (by setting a very close value of the required value of x). And no l'hopital's rule either if possible :-)

The limits are:

\[\lim_{x\rightarrow 0} \frac{ln(x^{2}+e^{x})}{ln(x^{4}+e^{2x})}\]

\[\lim_{x\rightarrow \infty } \frac{ln(x^{2}+e^{x})}{ln(x^{4}+e^{2x})}\]

Thank you.

 

[Yankel]

mmm...after rethinking the problem. If on the denominator, I take the power of 2 up, and then using ln rules takes it out of the ln, I get 0.5. Is this the answer for both cases ?

If so, I apologize, and here is a new question instead :-)

\[\lim_{x\rightarrow \infty }[(x+2)ln(x+2)-2(x+1)ln(x+1)+xln(x)]\]

 

[alyafey22]

mmm...after rethinking the problem. If on the denominator, I take the power of 2 up, and then using ln rules takes it out of the ln, I get 0.5. Is this the answer for both cases ?

If so, I apologize, and here is a new question instead :-)

\[\lim_{x\rightarrow \infty }[(x+2)ln(x+2)-2(x+1)ln(x+1)+xln(x)]\]

Hint: use that

$$\lim_{x \to \infty } x \log(x+1)-x \log(x) = 1 $$

 

[Yankel]

I'm afraid that the hint is not helping, I don't know how to proceed.

In addition, my solution to the first limits is wrong (final answer is correct, but the way is wrong).

 

[Greg]

The first one (using the limit definition of $e$ and approximation to Taylor series):

$$\lim_{x\to0}\dfrac{\ln(x^2+e^x)}{\ln(x^4+e^{2x})}=\lim_{x\to0}\dfrac{\ln(x^2+[(x+1)^{1/x}]^x)}{\ln(x^4+[(2x+1)^{1/x}]^x)}=\lim_{x\to0}\dfrac{\ln(x^2+x+1)}{\ln(x^4+2x+1)}\sim\dfrac{\ln(e^x)}{\ln(e^{2x})}=\dfrac12$$

Third one:

$$\ln\left(\lim_{x\to\infty}\dfrac{(x+2)^{x+2}\cdot x^x}{(x+1)^{2(x+1)}}\right)\sim\ln(1)=0$$

 

[Greg]

The first one (using the limit definition of $e$ and approximation to Taylor series):

$$\lim_{x\to0}\dfrac{\ln(x^2+e^x)}{\ln(x^4+e^{2x})}=\lim_{x\to0}\dfrac{\ln(x^2+[(x+1)^{1/x}]^x)}{\ln(x^4+[(2x+1)^{1/x}]^x)}=\lim_{x\to0}\dfrac{\ln(x^2+x+1)}{\ln(x^4+2x+1)}\sim\dfrac{\ln(e^x)}{\ln(e^{2x})}=\dfrac12$$

This technique (with the limit definition of $e$) is questionable. Consider using it to solve

$$\lim_{x\to0}\dfrac{e^x-x-1}{x^2}$$