# Properties - Examples

Example: [PolynomialLimit] Let $$p(x)=a_0+a_1x+\cdots+a_nx^n$$ be any polynomial with real coefficients. Show that $\lim_{x\to x_0} p(x)=p(x_0).$

We can see from [LimitProduct] that $\lim_{x\to x_0} x^2=\lim_{x\to x_0}x\cdot\lim_{x\to x_0}x=x_0\cdot x_0=x_0^2.$ We can now do this repeatedly to establish that $\lim_{x\to x_0} x^k=x_0^n$ for any positive integer $$k$$. We can then use [LimitConstantMultiple]to see that $\lim_{x\to x_0} a_kx^k=a_kx_0^n$ for any positive integer $$k$$. Finally we can apply [LimitAddition] repeatedly to show that the limit of sums of these terms is the sum of their limits. We conclude that $\lim_{x\to x_0} (a_0+a_1x+\cdots+a_nx^n)=a_0+a_1x_0+\cdots+a_nx_0^n.$

Example: [CosineLimit] Use the definition to show that $\lim_{x\to x_0} \cos x=\cos x_0.$

We did not specify the value of $$x_0$$, because it seems pretty obvious that $$\lim_{x\to x_0} \cos x = \cos x_0$$ for any $$x_0$$ in the reals. Let's prove that. Write $$x=x_0+h$$. For any $$\epsilon>0$$ we can use a trigonometric addition formula to write \begin{align*} \cos x &= \cos (x_0+h)\\ &= \cos x_0\cos h-\sin x_0\sin h\\ \end{align*} We now use [LimitAddition], [CosineZeroLimit], and [SineZeroLimit]  to see that \begin{align*} \lim_{x\to x_0}\cos x &= \cos x_0\lim_{h\to 0}\cos h-\sin x_0\lim_{h\to 0}\sin h\\ &= (\cos x_0)(1)-(\sin x_0)(0).\\ &= \cos x_0. \end{align*} Notice the bootstrap trick we ended up using here. We used some triangle inequalities with the definition to establish limits for the sine and cosine at 0, and then used trigonometric identities with those and the limit properties to get limits for all other values of $$x_0$$.

Example: [CosDiffLimit] Show that $\lim_{x\to0} \frac{\cos x -1}{x^2}=-\frac12.$

We wanted to use a trigonometric identity in [SineLimit]; this time we have a good trigonometric limit to work with, so we can. We'll multiply the quantity whose limit we want to find by 1: \begin{align*} \frac{\cos x -1}{x^2} &= \frac{\cos x -1}{x^2}\frac{\cos x +1}{\cos x +1}\\ &= \frac{\cos^2 x-1}{x^2(\cos x+1)}\\ &= \frac{-\sin^2 x}{x^2(\cos x+1)}\\ &= -\left(\frac{\sin x}{x}\right)^2\frac{1}{\cos x+1}. \end{align*} But now we can apply [SineLimit], [LimitProduct], and [LimitReciprocal]to get $\lim_{x\to0} \frac{\cos x -1}{x^2}=-(1)^2(\frac12)=-\frac12$

We can now easily show the following somewhat surprising result.
Example: [CosFirstOrderLimit] Show that $\lim_{x\to0} \frac{\cos x -1}{x}=0.$

Using [LimitProduct] and [CosDiffLimit] we get immediately that $\lim_{x\to0} \frac{\cos x -1}{x}=\lim_{x\to0} \frac{x(\cos x -1)}{x^2}= \lim_{x\to0}x\cdot\lim_{x\to0}\frac{\cos x -1}{x^2}=0\cdot\frac{-1}2=0 .$