Limit of difference quotient of $\frac{t^4}{2}+t^3-\frac{t^2}{2}+1$

I am trying to compute the limit of the difference quotient of the function $x(t)=\frac{t^4}{2}+t^3-\frac{t^2}{2}+1$ at $t=1$, i.e. $$\lim_{\ \Delta t\to 0}\frac{x(1+\Delta t)-x(1)}{\Delta t}$$.

Since $x(1+\Delta t)=\frac{\Delta t^4+6\Delta t^3+11\Delta t^2+8\Delta t+4}{2}$ and $x(1)=2$ we have $$\lim_{\ \Delta t\to 0}\frac{x(1+\Delta t)-x(1)}{\Delta t}=\lim_{\Delta t\to 0}\frac{\frac{\Delta t^4+6\Delta t^3+11\Delta t^2+8\Delta t+4}{2}-1}{\Delta t}=\frac{1}{2}\lim_{\Delta t\to 0}\frac{\Delta t^4+6\Delta t^3+11\Delta t^2+8\Delta t+2}{\Delta t}$$.

Now, I have tried factorizing the polynomial, changing variables, etc. but I haven't been able to obtain the value $4$ (which I know is true by differentiating the polynomial and setting $t=1$) and worse, I haven't been able to not make it diverge to infinity.

So, I would appreciate if someone could tell me how to handle this limit, thanks.

Solution 1:

You made a mistake in that step, the correct is as below $$\lim_{\ \Delta t\to 0}\frac{x(1+\Delta t)-x(1)}{\Delta t}=\lim_{\Delta t\to 0}\frac{\frac{\Delta t^4+6\Delta t^3+11\Delta t^2+8\Delta t}{2}}{\Delta t}=\frac{1}{2}\lim_{\Delta t\to 0}\frac{\Delta t^3+6\Delta t^2+11\Delta t+8}{2}=4$$.