# Specific Point of Limit on graph

• Aug 5th 2009, 08:59 AM
skeske1234
Specific Point of Limit on graph
Interpret the limit as the slope of the tangent line to a curve at a specific point after you evaluate the limit.

lim[(1+h)^(1/6)-1]/h
h->0

I evaluated the limit to be
M_t=1/6 to y=x^(1/6) but I am having trouble finding the x co-ordinate.

The answer in the back says:
M_t=1/6 to y=x^(1/6) at x=1

how did they get x=1?
• Aug 5th 2009, 09:09 AM
Plato
Quote:

Originally Posted by skeske1234
Interpret the limit as the slope of the tangent line to a curve at a specific point after you evaluate the limit.
lim[(1+h)^(1/6)-1]/h
h->0
I evaluated the limit to be
M_t=1/6 to y=x^(1/6) but I am having trouble finding the x co-ordinate.
The answer in the back says:
M_t=1/6 to y=x^(1/6) at x=1
how did they get x=1?

Actually you are given that $\displaystyle \color{red}x=1$.
It is part of the statement of the problem. See above.
• Aug 5th 2009, 01:26 PM
DeMath
Quote:

Originally Posted by skeske1234
Interpret the limit as the slope of the tangent line to a curve at a specific point after you evaluate the limit.

lim[(1+h)^(1/6)-1]/h
h->0

I evaluated the limit to be
M_t=1/6 to y=x^(1/6) but I am having trouble finding the x co-ordinate.

The answer in the back says:
M_t=1/6 to y=x^(1/6) at x=1

how did they get x=1?

If you need evaluate the $\displaystyle \mathop {\lim }\limits_{h \to 0} \frac{{{{\left( {1 + h} \right)}^{1/6}} - 1}}{h}$ then, using the formula for the difference of cubes, we get

$\displaystyle {\left( {1 + h} \right)^{1/2}} - 1 = \left( {{{\left( {1 + h} \right)}^{1/6}} - 1} \right)\left( {{{\left( {1 + h} \right)}^{1/3}} + {{\left( {1 + h} \right)}^{1/6}} + 1} \right) \Leftrightarrow$

$\displaystyle \Leftrightarrow {\left( {1 + h} \right)^{1/6}} - 1 = \frac{{{{\left( {1 + h} \right)}^{1/2}} - 1}}{{{{\left( {1 + h} \right)}^{1/3}} + {{\left( {1 + h} \right)}^{1/6}} + 1}}.$

So we have

$\displaystyle \mathop {\lim }\limits_{h \to 0} \frac{{{{\left( {1 + h} \right)}^{1/6}} - 1}}{h} = \mathop {\lim }\limits_{h \to 0} \frac{{{{\left( {1 + h} \right)}^{1/2}} - 1}}{{h\left( {{{\left( {1 + h} \right)}^{1/3}} + {{\left( {1 + h} \right)}^{1/6}} + 1} \right)}} =$

$\displaystyle = \mathop {\lim }\limits_{h \to 0} \frac{{\left( {{{\left( {1 + h} \right)}^{1/2}} - 1} \right)\left( {{{\left( {1 + h} \right)}^{1/2}} + 1} \right)}}{{h\left( {{{\left( {1 + h} \right)}^{1/3}} + {{\left( {1 + h} \right)}^{1/6}} + 1} \right)\left( {{{\left( {1 + h} \right)}^{1/2}} + 1} \right)}} =$

$\displaystyle = \mathop {\lim }\limits_{h \to 0} \frac{1}{{\left( {{{\left( {1 + h} \right)}^{1/3}} + {{\left( {1 + h} \right)}^{1/6}} + 1} \right)\left( {{{\left( {1 + h} \right)}^{1/2}} + 1} \right)}} = \frac{1}{{3 \cdot 2}} = \frac{1}{6}.$
• Aug 5th 2009, 01:32 PM
Krizalid
Put $\displaystyle t=\sqrt[6]{h+1}$ so that $\displaystyle t\to1$ and $\displaystyle h=t^6-1.$