Lesson

There are a number of limit laws which we tend to use in limit questions without writing them into our calculations in a formal manner.

1. The limit of a sum or difference is the same as the sum or difference of the limits of each part.

$\lim_{x\rightarrow a}f\left(x\right)\pm g\left(x\right)=\lim_{x\rightarrow a}f\left(x\right)\pm\lim_{x\rightarrow a}g\left(x\right)$lim`x`→`a``f`(`x`)±`g`(`x`)=lim`x`→`a``f`(`x`)±lim`x`→`a``g`(`x`)

2. The limit of a constant multiple of a function is equal to a constant multiple of the limit of a function.

$\lim_{x\rightarrow a}\left(k\cdot f\left(x\right)\right)=k\cdot\lim_{x\rightarrow a}f\left(x\right)$lim`x`→`a`(`k`·`f`(`x`))=`k`·lim`x`→`a``f`(`x`)

3. The limit of a product/quotient is the same as the product/quotient of the limits of each function.

$\lim_{x\rightarrow a}\frac{f\left(x\right)}{g\left(x\right)}=\frac{\lim_{x\rightarrow a}f\left(x\right)}{\lim_{x\rightarrow a}g\left(x\right)}$lim`x`→`a``f`(`x`)`g`(`x`)=lim`x`→`a``f`(`x`)lim`x`→`a``g`(`x`)

4. For the constant function,$f\left(x\right)=k,\lim_{x\rightarrow a}f\left(x\right)=k$`f`(`x`)=`k`,lim`x`→`a``f`(`x`)=`k`

For well-behaved functions that don't have jumps or breaks near where we're finding the limit, finding a limit at a point in the domain is usually a matter of evaluating the function at that point.

So for a function $f\left(x\right)$`f`(`x`) that has no jumps around $x=a$`x`=`a`:

$\lim_{x\to a}f\left(x\right)=f\left(a\right)$lim`x`→`a``f`(`x`)=`f`(`a`)

Other functions may have points where they are undefined, which can mean a limit does not exist at those points, or it can happen that a limit does exist at such points and we need to use the limit theorems given above combined with knowledge of previously discovered limits to evaluate it.

In the chapter on the differentiation of the sine function, we found that we needed to evaluate the limit $\lim_{\theta\rightarrow0}\frac{\sin\theta}{\theta}$lim`θ`→0`s``i``n``θ``θ`. If $0$0 is substituted for $\theta$`θ` we obtain the indeterminate expression $\frac{0}{0}$00 and so, some further reasoning was needed to show that the limit is $1$1 in this case.

We also needed to evaluate $\lim_{\theta\rightarrow0}\frac{\cos\theta-1}{\theta}$lim`θ`→0`c``o``s``θ`−1`θ` which, again, leads to the indeterminate expression $\frac{0}{0}$00 if we simply substitute $0$0 for $\theta$`θ`. Some algebraic manipulation was needed and the third of the above limit theorems was applied to show that the limit, in this case, is $0$0.

Finally, these two facts were combined using the first of the above limit theorems to deduce that the derivative of the sine function is the cosine function.

Evaluate $\lim_{y\rightarrow25}\log_2\left(59+\sqrt{y}\right)$lim`y`→25`l``o``g`2(59+√`y`).

The function given by $\log_2\left(59+\sqrt{y}\right)$`l``o``g`2(59+√`y`) is defined everywhere on the domain $y>0$`y`>0 with no jumps. So, we can substitute $25$25 for $y$`y` and simplify to obtain the limiting value $\log_264$`l``o``g`264, which is $6$6.

Evaluate $\lim_{t\rightarrow0}\frac{\sin^2t+t^3}{t^2}$lim`t`→0`s``i``n`2`t`+`t`3`t`2.

Substitution of $t=0$`t`=0 gives the indeterminate form $\frac{0}{0}$00. So, we use the limit theorems. Using theorem $1$1, we write the statement as $\lim_{t\rightarrow0}\frac{\sin^2t}{t^2}+\lim_{t\rightarrow0}\frac{t^3}{t^2}$lim`t`→0`s``i``n`2`t``t`2+lim`t`→0`t`3`t`2.

We split this further with the help of the third theorem.

$\lim_{t\rightarrow0}\frac{\sin t}{t}.\lim_{t\rightarrow0}\frac{\sin t}{t}+\lim_{t\rightarrow0}\frac{t^3}{t^2}$lim`t`→0`s``i``n``t``t`.lim`t`→0`s``i``n``t``t`+lim`t`→0`t`3`t`2.

Now, we evaluate the separate limit statements. For $\lim_{t\rightarrow0}\frac{t^3}{t^2}$lim`t`→0`t`3`t`2 we note that when $t$`t` is very near but not equal to $0$0, the statement is equivalent to $\lim_{t\rightarrow0}\ t$lim`t`→0 `t` which is $0$0. So, we have

$\lim_{t\rightarrow0}\frac{\sin^2t+t^3}{t^2}=1\times1+0=1$lim`t`→0`s``i``n`2`t`+`t`3`t`2=1×1+0=1.

Find the limit, $\lim_{x\rightarrow0}\frac{\tan x}{3x}$lim`x`→0`t``a``n``x`3`x`.

This can be re-written as $\lim_{x\rightarrow0}\frac{1}{3}\frac{\sin x}{x}\times\lim_{x\rightarrow0}\frac{1}{\cos x}=\frac{1}{3}\times1\times1=\frac{1}{3}$lim`x`→013`s``i``n``x``x`×lim`x`→01`c``o``s``x`=13×1×1=13

If $\lim_{x\to3}f\left(x\right)=7$lim`x`→3`f`(`x`)=7, find $\lim_{x\to3}\left(1+f\left(x\right)\right)^2$lim`x`→3(1+`f`(`x`))2.

Find $\lim_{x\to0}\left(\frac{\tan x}{x}\right)$lim`x`→0(`t``a``n``x``x`).

Identify discontinuities and limits of functions

Apply differentiation methods in solving problems