Calculate limits

[JasaMath]

Hi, I would kindly ask for help calculating this two limits. As you can see on the picture I managed to do the first step in both of them, but I don't know how to continue on the first one. I know that I should use L'Hopital rule on the second one as I did but I ended being stuck.

Thanks for help!
Jasa

 

[Deleted member 4993]

Hi, I would kindly ask for help calculating this two limits. As you can see on the picture I managed to do the first step in both of them, but I don't know how to continue on the first one. I know that I should use L'Hopital rule on the second one as I did but I ended being stuck. Thanks for help!
Jasa

Your attachment is extremely faint - and I cannot read it.

Please re-write it with black ink and ample white space!!

 

[Steven G]

Your very 1st equal sign in the 2nd problem is not valid. Whenever you compute the derivative of any trig function the angle never changes. Again, whenever you compute the derivative of any trig function the any angle never changes. Is this clear?

 

[Steven G]

sqrt(2^n) grows much much quicker than n^5----same as 2^n grows much much faster than n^10

 

[JasaMath]

Your very 1st equal sign in the 2nd problem is not valid. Whenever you compute the derivative of any trig function the angle never changes. Again, whenever you compute the derivative of any trig function the any angle never changes. Is this clear?

Oh, I see that mistake now, thank you. It is very clear.

 

[JasaMath]

sqrt(2^n) grows much much quicker than n^5----same as 2^n grows much much faster than n^10

So I should just divide it by sqrt(2^n)?

 

[skeeter]

for the radical limit ... after rationalizing, recommend multiplying numerator & denominator by [MATH]\dfrac{1}{\sqrt{2^n}}[/MATH]
the trig limit can be found with two iterations of L'Hopital

 

[pka]

Hi, I would kindly ask for help calculating this two limits. As you can see on the picture I managed to do the first step in both of them, but I don't know how to continue on the first one. I know that I should use L'Hopital rule on the second one as I did but I ended being stuck.

The first limit is a real beast to tame.
It can be written as \(\mathop {\lim }\limits_{n \to \infty } \frac{{{n^5} - {n^3}}}{{\sqrt {{2^n} + {n^5}} + \sqrt {{2^n} + {n^3}} }}\) It has limit zero, SEE HERE

 

[JasaMath]

for the radical limit ... after rationalizing, recommend multiplying numerator & denominator by [MATH]\dfrac{1}{\sqrt{2^n}}[/MATH]
the trig limit can be found with two iterations of L'Hopital

Thank you, that is what I needed!

 

[skeeter]

[MATH]\lim_{n \to \infty} \dfrac{n^5 - n^3}{\sqrt{2^n+n^5} + \sqrt{2^n+n^3}} \cdot \dfrac{\frac{1}{\sqrt{2^n}}}{\frac{1}{\sqrt{2^n}}}[/MATH]
[MATH]\lim_{n \to \infty} \dfrac{\frac{n^5}{2^{n/2}} - \frac{n^3}{2^{n/2}}}{\sqrt{1+\frac{n^5}{2^n}} + \sqrt{1+\frac{n^3}{2^n}}}[/MATH]
the exponential term in each fraction's denominator is dominant ... as [MATH]n \to \infty[/MATH], all fractional terms go to zero.

 

[skeeter]

You can also find the trig limit using basic identities ...

[MATH]\dfrac{1-\cos(2x) + \tan^2{x}}{x\sin{x}}[/MATH]
[MATH]\dfrac{1 - (1-2\sin^2{x}) + \frac{\sin^2{x}}{\cos^2{x}}}{x\sin{x}}[/MATH]
[MATH]\dfrac{\sin^2{x}}{x\sin{x}} \cdot (2+\sec^2{x})[/MATH]
[MATH]\lim_{x \to 0} \dfrac{\sin{x}}{x} \cdot (2+\sec^2{x})[/MATH]