Therefore, the limit is equal to $\boxed{2\ln 2-2+\pi}$.

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In summary, the limit $\boxed{2\ln 2-2+\pi}$ represents the value that a function approaches as its input approaches a specific value or as the number of terms in a sequence increases without bound. It is calculated by evaluating the function or sequence at the specific value or as the number of terms increases without bound, and then simplifying the resulting expression. The limit $\boxed{2\ln 2-2+\pi}$ will have the same value at all points where the function or sequence approaches the specific value or as the number of terms increases without bound. However, it may be infinite if the function or sequence does not approach a specific value or as the number of terms increases without bound. In mathematics, the
  • #1
Chris L T521
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Thanks again to those who participated in last week's POTW! Here's this week's problem!

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Problem: Evaluate $\displaystyle\lim_{x\to 0^{+}}\left(\int_{x}^{1}\dfrac{\ln{(1+t)}}{\sqrt{t}}dt+\int_{0}^{x}\dfrac{\sin{2t}}{\sqrt{4+t^2}\int_{0}^{x}(\sqrt{y+1}-1)dy}dt\right)$.

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Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!
 
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This week's problem was correctly answered by Opalg and Pranav. You can find Pranav's solution below.

[sp]The given limit can be written as:
$$\int_0^1 \frac{\ln(1+t)}{\sqrt{t}}\,dt + \lim_{x\rightarrow 0^+} \cfrac{\int_0^x \frac{\sin(2t)}{\sqrt{4+t^2}}\,dt}{\int_0^x (\sqrt{y+1}-1)\,dy}$$
I evaluate the above separately.

First I evaluate the following integral:
$$\int_0^1 \frac{\ln(1+x)}{\sqrt{t}}\,dt$$
Substitute $t=x^2 \Rightarrow dt=2x\,dx$, the integral becomes
$$2\int_0^1 \ln(1+x^2)\,dx=2\left(\left(\ln(1+x^2)\cdot x\right|_0^1-\int_0^1 \frac{2x^2}{1+x^2}\,dx\right)$$
(I used integration by parts)
$$=2\left(\ln 2-2\left(\int_0^1 1-\frac{1}{1+x^2} \,\,dx\right)\right)=2\left(\ln 2 -2\left(1-\frac{\pi}{4}\right)\right)=\boxed{2\ln 2-4+\pi}$$

Next, I evaluate the following limit,
$$\lim_{x\rightarrow 0^+} \cfrac{\int_0^x \frac{\sin(2t)}{\sqrt{4+t^2}}\,dt}{\int_0^x (\sqrt{y+1}-1)\,dy}$$
Substituting $x=0$ gives $\frac{0}{0}$ so L'Hopital's rule can be used i.e
$$\lim_{x\rightarrow 0^+} \cfrac{\int_0^x \frac{\sin(2t)}{\sqrt{4+t^2}}\,dt}{\int_0^x (\sqrt{y+1}-1)\,dy}=\lim_{x\rightarrow 0^+} \cfrac{\frac{\sin 2x}{\sqrt{4+x^2}}}{\sqrt{x+1}-1}=\lim_{x\rightarrow 0^+} \frac{\sin 2x}{\sqrt{4+x^2}(\sqrt{x+1}-1)}=\lim_{x\rightarrow 0^+} \frac{\sin (2x) (\sqrt{x+1}+1)}{x\sqrt{4+x^2}}$$
$$=\lim_{x\rightarrow 0^+} 2\frac{\sin (2x) (\sqrt{x+1}+1)}{2x\sqrt{4+x^2}}$$
The limit of $\sin(2x)/(2x)$ as $x$ tends to zero is 1, hence, the above limit is equal to $\boxed{2}$.

Putting everything together, the final result is:
$$2\ln 2-4+\pi+2=\boxed{2\ln 2-2+\pi}$$[/sp]
 

FAQ: Therefore, the limit is equal to $\boxed{2\ln 2-2+\pi}$.

What does the limit $\boxed{2\ln 2-2+\pi}$ represent?

The limit $\boxed{2\ln 2-2+\pi}$ represents the value that a function approaches as its input approaches a specific value or as the number of terms in a sequence increases without bound.

How is the limit $\boxed{2\ln 2-2+\pi}$ calculated?

The limit $\boxed{2\ln 2-2+\pi}$ is calculated by evaluating the function or sequence at the specific value or as the number of terms increases without bound, and then simplifying the resulting expression.

Can the limit $\boxed{2\ln 2-2+\pi}$ have different values at different points?

No, the limit $\boxed{2\ln 2-2+\pi}$ will have the same value at all points where the function or sequence approaches the specific value or as the number of terms increases without bound.

Is the limit $\boxed{2\ln 2-2+\pi}$ always a finite value?

No, the limit $\boxed{2\ln 2-2+\pi}$ may be infinite if the function or sequence does not approach a specific value or as the number of terms increases without bound.

How is the limit $\boxed{2\ln 2-2+\pi}$ used in mathematics?

The limit $\boxed{2\ln 2-2+\pi}$ is used to understand and analyze the behavior of functions and sequences, specifically at specific points or as the number of terms increases without bound. It is also used to determine if a function or sequence is continuous at a specific point or if it has a limit at that point.

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