# Limit...using given info

• Dec 7th 2007, 04:52 AM
polymerase
Limit...using given info
Given that the tangent line to the graph of $\displaystyle f(x)$ at (0,0) has equation $\displaystyle 2y=x$ and that $\displaystyle f(x)$ has a horizontal asymptote at $\displaystyle \infty$ with equation $\displaystyle y=2$, find the value of

$\displaystyle \displaystyle\lim_{x\to{0^{+}}} {\frac{sin(2x)}{f(x)}} - x^2f(\frac{1}{x})$
• Dec 7th 2007, 05:35 AM
topsquark
Quote:

Originally Posted by polymerase
Given that the tangent line to the graph of $\displaystyle f(x)$ at (0,0) has equation $\displaystyle 2y=x$ and that $\displaystyle f(x)$ has a horizontal asymptote at $\displaystyle \infty$ with equation $\displaystyle y=2$, find the value of

$\displaystyle \displaystyle\lim_{x\to{0^{+}}} {\frac{sin(2x)}{f(x)}} - x^2f(\frac{1}{x})$

I'm not any sort of an expert at limits, but this looks (Cool) so I'm going to take a shot at it.
First since $\displaystyle \lim_{x \to +\infty}f(x) = 2$ we know that $\displaystyle \lim_{x \to 0^+}f \left ( \frac{1}{x} \right ) = 2$

Thus
$\displaystyle \lim_{x \to 0+}x^2 f \left ( \frac{1}{x} \right ) = \lim_{x \to 0^+} 2x^2 = 0$

Next we have $\displaystyle \lim_{x \to 0^+} \frac{sin(2x)}{f(x)}$. We know that $\displaystyle \lim_{x \to 0}f(x) = 0$ since the tangent line to y = f(x) at x = 0 is 2y = x. Thus the limit is of the form $\displaystyle \frac{0}{0}$ and we may apply L'Hopital's rule:
$\displaystyle \lim_{x \to 0^+} \frac{sin(2x)}{f(x)} = \lim_{x \to 0^+} \frac{2 cos(2x)}{\frac{1}{2}} = \lim_{x \to 0^+}4cos(2x) = 4$
(Because $\displaystyle f(x) \approx \frac{1}{2}x$ near x = 0.)

Thus
$\displaystyle \lim_{x \to 0^+} { \frac{sin(2x)}{f(x)}} - x^2f(\frac{1}{x}) = 4 - 0 = 4$

-Dan
• Dec 7th 2007, 05:53 AM
polymerase
Quote:

Originally Posted by topsquark
I'm not any sort of an expert at limits, but this looks (Cool) so I'm going to take a shot at it.
First since $\displaystyle \lim_{x \to +\infty}f(x) = 2$ we know that $\displaystyle \lim_{x \to 0^+}f \left ( \frac{1}{x} \right ) = 2$

Thus
$\displaystyle \lim_{x \to 0+}x^2 f \left ( \frac{1}{x} \right ) = \lim_{x \to 0^+} 2x^2 = 0$

Next we have $\displaystyle \lim_{x \to 0^+} \frac{sin(2x)}{f(x)}$. We know that $\displaystyle \lim_{x \to 0}f(x) = 0$ since the tangent line to y = f(x) at x = 0 is 2y = x. Thus the limit is of the form $\displaystyle \frac{0}{0}$ and we may apply L'Hopital's rule:
$\displaystyle \lim_{x \to 0^+} \frac{sin(2x)}{f(x)} = \lim_{x \to 0^+} \frac{2 cos(2x)}{\frac{1}{2}} = \lim_{x \to 0^+}4cos(2x) = 4$
(Because $\displaystyle f(x) \approx \frac{1}{2}x$ near x = 0.)

Thus
$\displaystyle \lim_{x \to 0^+} { \frac{sin(2x)}{f(x)}} - x^2f(\frac{1}{x}) = 4 - 0 = 4$

-Dan

Thanks!
By the way, you're right :).
• Dec 7th 2007, 06:17 AM
topsquark
Quote:

Originally Posted by polymerase
Thanks!
By the way, you're right :).

(Whew)

That was fun! Thank you for the problem.

-Dan
• Dec 7th 2007, 06:41 AM
polymerase
Quote:

Originally Posted by topsquark
(Whew)

That was fun! Thank you for the problem.

-Dan

You think you had a hard time...i had to do this problem for my calculus term test.:eek: