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tmacmvp13579 · May 13, 2008, 07:39 PM

1.If
f(x)=− 25/ (2x+3)
find

lim ( f(1+h)−f(1)) /h
h→ 0

2.lim (2t^2−3t−2)/(t^2+t-6)
t→ 2

3.lim (x^2+4x-5)/(x^2+x-2)
x→ 1

4.lim x^2 / |x|
x→ 0

5.lim 1/|x|
x→ 0

galactus · May 14, 2008, 06:36 AM

1.If
f(x)=− 25/ (2x+3)
find

lim ( f(1+h)−f(1)) /h
h→ 0

Plug in 1+h for x in your expression and reduce.

2.lim (2t^2−3t−2)/(t^2+t-6)
t→ 2

Long divide first. This is an indeterminate form.

3.lim (x^2+4x-5)/(x^2+x-2)
x→ 1

Same as #2

.lim x^2 / |x|
x→ 0

This is $\frac{x^{2}}{x}=x$ or $\frac{x^{2}}{-x}=-x$

5.lim 1/|x|
x→ 0

Think of it as $\frac{1}{\sqrt{x^{2}}}$

Look at the graph.