Evaluate $\lim_{h\to 0}\frac 1 h\int_{-\infty}^\infty g\left(\frac x h\right)f(x)\,\mathrm dx$

Simply substituting $u=x/h$ gives $$ \frac{1}{h} \int_{-\infty}^{\infty} g\left(\frac{x}{h}\right)f(x)\,dx = \int_{-\infty}^{\infty} g(u) f(hu) \,du = \int_{-1}^{1} g(u) f(hu) \,du $$ where the latter equality follows from the definition of $g$.

Given that both $f$ and $g$ are continuous on the closed interval $[-1,1]$, they are uniformly continuous there, so we may pull the limit under the integral: $$ \lim_{h\to 0} \frac{1}{h} \int_{-\infty}^{\infty} g\left(\frac{x}{h}\right)f(x)\,dx = \lim_{h\to 0} \int_{-1}^{1} g(u) f(hu) \,du = \int_{-1}^{1} g(u) \cdot\lim_{h\to 0}f(hu) \,du$$

Finally, by continuity, $\lim_{h\to 0}f(hu) = f(0)$ and $$ \lim_{h\to 0} \frac{1}{h} \int_{-\infty}^{\infty} g\left(\frac{x}{h}\right)f(x)\,dx = f(0)\int_{-1}^{1}g(u) du = f(0)$$

Since $g\left(x\right)=0 $ if $\left|x\right|\geq1 $ we have $$I_{h}=\int_{-\infty}^{\infty}g\left(\frac{x}{h}\right)f\left(x\right)dx=\int_{-h}^{h}g\left(\frac{x}{h}\right)f\left(x\right)dx $$ and now using the mean value theorem for integrals we have that exists some $c_{h}\in\left[-h,h\right] $ such that $$I_{h}=f\left(c_{h}\right)\int_{-h}^{h}g\left(\frac{x}{h}\right)dx=hf\left(c_{h}\right)\int_{-1}^{1}g\left(u\right)du=hf\left(c_{h}\right) $$ hence $$\lim_{h\rightarrow0}\frac{I_{h}}{h}=\lim_{h\rightarrow0}f\left(c_{h}\right)=\color{red}{f\left(0\right)}$$ from the squeeze theorem and using the fact that $f$ is continuous.