Sandwich Theorem

[Fuzell Wood]

Again I don't know what to do,but there is a hint in the test that i should use Squeeze Theorem,but still i haven't managed to solve it.

 

[MarkFL]

I could be way off base here, but what I would do is differentiate both sides of the given equation with respect to $t$, to obtain:

[MATH]\frac{f^4(t)}{f^2(t)+1}f'(t)-\frac{t^4}{t^2+1}=0[/MATH]
Hence:

[MATH]f'(t)=\frac{t^4(f^2(t)+1)}{(t^2+1)f^4(t)}[/MATH]
Now, since:

[MATH]\lim_{x\to\infty}\left(\frac{d}{dx}\left(\frac{x^4}{x^2+1}\right)\right)=\infty[/MATH]
we can say:

[MATH]\lim_{t\to\infty}f(t)=t[/MATH]
Hence:

[MATH]\lim_{x\to\infty}f'(x)=1[/MATH]

 

[pka]

Again I don't know what to do,but there is a hint in the test that i should use Squeeze Theorem,but still i haven't managed to solve it.

View attachment 12235

This is a comment on the question itself. If each of $\displaystyle \alpha~\&~\beta$ is a differentiable function and $\displaystyle F(x) = \int_{\alpha (x)}^{\beta (x)} {g(t)dt}$ then the derivative $\displaystyle F'(x) = g(\beta (x))\beta '(x) - g(\alpha (x))\alpha '(x)$
Now while I was writing this MarkFL made is post, with most of which I agree. But the more I thought about it I wonder: what can it mean that $\displaystyle f(t)$ is a number such that an integral has a value 2? How is that a differential function? Now it may be, but recall that this is supposedly a university entry exam. Therefore, I think the question is simply flawed.

 

[LCKurtz]

I think it's pretty clear that [MATH]f(t)\ge t[/MATH] and [MATH]f(t)-t[/MATH] decreases to [MATH]0[/MATH] so [MATH]f'(t) \le 1[/MATH]. But I don't think that is enough to say [MATH]f'(t) \to 1[/MATH] as [MATH]t\to \infty[/MATH].