# Math Help - Calculus 6e James Stewart

1. ## Calculus 6e James Stewart

Calculus 6th Edition
Metric International Edition
James Stewart

page 86.

55)
If $\lim_{x\rightarrow 1}\frac{f(x)-8}{x-1}=10$, find $\lim_{x\rightarrow 1}f(x)$.

57)
If $f(x)=x^2$ if $x$ is rational.
$=0$ if $x$ is irrational.

prove that $\lim_{x\rightarrow 0}f(x)=0$.

59)
Show by means of an example that $\lim_{x\rightarrow a}[f(x)g(x)]$ may exist even though neither $\lim_{x\rightarrow a}f(x)$ nor $\lim_{x\rightarrow a}g(x)$ exists.

61)
Is there a number $a$ such that

$\lim_{x\rightarrow -2}\frac{3x^2+ax+a+3}{x^2+x-2}$

exist? If so, find the value of $a$ and the value of the limit.

2. The first one should be 8 since you want 0/0 here.

I would use the squeeze/turkey sandwich theorem here.
f is either the square or 0. Hence

$0\le |f(x)-0|\le x^2\to 0$ as $x\to 0$.

Third one... $g(x)=f(x)=(-1)^x$ so $g(x)f(x)=1$ for all x.
Here's I'm thinking of x as a natural number heading towards infinity, but I don't think it matters where we head.

Iin the last one we need
$3(-2)^2+a(-2)+a+3=0$, which seems to be 15.
It's 2am and I'm done for tonight.

With a=2 we have

$\lim_{x\to -2}{3(x^2+5x+6)\over x^2+x-2}=\lim_{x\to -2}{3(x+3)(x+2)\over (x+2)(x-1)}$

$=3\lim_{x\to -2}{x+3\over x-1}$..............

3. Originally Posted by matheagle
The first one should be 8 since you wnat 0/0 here.
Why I should make it becomes 0/0?

Originally Posted by matheagle
Third one... $g(x)=f(x)=(-1)^x$ so $g(x)f(x)=1$ for all x.
Here's I'm thinking of x as a natural number heading towards infinity, but I don't think it matters where we head.
Is it means for all x in the domain(Natural numbers)? If yes, the function will be discontinuous, then how can it be differentiated?

Originally Posted by matheagle
the last one we need
$3(-2)^2+a(-2)+a+3=0$, which seems to be 15.
It's 2am and I'm done for tonight.

With a=2 we have

$\lim_{x\to -2}{3(x^2+5x+6)\over x^2+x-2}=\lim_{x\to -2}{3(x+3)(x+2)\over (x+2)(x-1)}$

$=3\lim_{x\to -2}{x+3\over x-1}$..............
Again, why should I make it 0/0?

4. The denomiator is heading towards zero.
If you want the limit to be 8 the numerator must also be heading towards zero.
Same comment on the last one, if the numerator is heading towards, say 5, and the denominator
is heading towards zero, then you either have a plus or minus infinity, or a does not exist situation.