# Math Help - Limit...using given info

1. ## Limit...using given info

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

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

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

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

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

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

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

-Dan

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

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

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

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

-Dan
Thanks!
By the way, you're right .

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

That was fun! Thank you for the problem.

-Dan

5. Originally Posted by topsquark

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.