# Let r(t) = \left \langle t^{-2}, \sin t, 7 \right \rangle|. Evaluate the limit: \lim_{h \to 0}...

## Question:

Let {eq}\displaystyle r(t) = \left \langle t^{-2}, \sin t, 7 \right \rangle {/eq}|.

Evaluate the limit:

{eq}\displaystyle \lim_{h \to 0} \frac{r(t + h) - r(t)}{h} {/eq}

## Limit:

Given: {eq}r(t) = \left \langle t^{-2}, \sin t, 7 \right \rangle| {/eq}

To find: {eq}\displaystyle \lim_{h \to 0} \frac{r(t + h) - r(t)}{h} {/eq}

In order to find {eq}r(t+h), \text{ replace } t \text{ by } t + h {/eq} in {eq}r(t) = \left \langle t^{-2}, \sin t, 7 \right \rangle {/eq}

{eq}r(t) = \left \langle t^{-2}, \sin t, 7 \right \rangle {/eq}

{eq}r(t+h) = \left \langle (t+h)^{-2}, \sin (t+h), 7 \right \rangle {/eq}

So,

{eq}r(t+h)-r(t)=\left \langle (t+h)^{-2}, \sin (t+h), 7 \right \rangle-\left \langle t^{-2}, \sin t, 7 \right \rangle\\ =\left \langle (t+h)^{-2}-t^{-2}, \sin (t+h)- \sin t,7-7\right \rangle\\ =\left \langle \frac{1}{(t+h)^2}-\frac{1}{t^2}, \sin (t+h)- \sin t,0 \right \rangle\\ =\left \langle \frac{t^2-(t+h)^2}{t^2(t+h)^2}, \sin (t+h)- \sin t,0 \right \rangle\\ =\left \langle \frac{-h^2-2ht}{t^2(t+h)^2}, \sin (t+h)- \sin t,0 \right \rangle {/eq}

Therefore,

{eq}\frac{r(t+h)-r(t)}{h}=\left \langle \frac{-h^2-2ht}{ht^2(t+h)^2}, \frac{\sin (t+h)- \sin t}{h},0 \right \rangle\\ \lim_{h\rightarrow 0}\frac{r(t+h)-r(t)}{h}=\lim_{h\rightarrow 0}\left \langle \frac{-h^2-2ht}{ht^2(t+h)^2}, \frac{\sin (t+h)- \sin t}{h},0 \right \rangle\\ =\lim_{h\rightarrow 0}\left \langle \frac{-h-2t}{t^2(t+h)^2}, \frac{\sin (t+h)- \sin t}{h},0 \right \rangle \text{ (using limit definition of derivative of } \sin t)\\ =\left \langle \frac{-2}{t^3},1-\cos t,0 \right \rangle {/eq}