I see the image contains three limit problems: 11, 13, and 15. Let's solve each one step by step.

Problem 11:

$\lim_{{x \to 0^+}} \frac{{\sin(4x)}}{{5x}}$ Here, we can use the limit property: $\lim_{x \to 0} \frac{\sin(kx)}{x} = k$, which holds true because $\frac{\sin(4x)}{x}$ approaches $4$ as $x \to 0$.

Thus: $\lim_{{x \to 0^+}} \frac{{\sin(4x)}}{{5x}} = \frac{4}{5}$

Problem 13:

$\lim_{{x \to \infty}} \frac{{2x^3 - 7x^2}}{{4x^3 + x + 9}}$ When taking the limit as $x \to \infty$, we divide both the numerator and denominator by the highest power of $x$, which is $x^3$: $\lim_{{x \to \infty}} \frac{{2x^3 - 7x^2}}{{4x^3 + x + 9}} = \lim_{{x \to \infty}} \frac{{2 - \frac{7}{x}}}{{4 + \frac{1}{x^2} + \frac{9}{x^3}}}$ As $x$ approaches infinity, the terms with $x$ in the denominator will approach zero: $= \frac{2}{4} = \frac{1}{2}$

Problem 15:

$\lim_{{x \to 3}} f(x)$ where