# Limits Problems

• Apr 5th 2011, 12:51 AM
Limits Problems
Hi. I need help. My test is next week, and on it there will be a lot of math stuff including limits. Now, the teacher gave us a few problems to show us what to expect on the test. That was last week. I've been trying to solve these limits:

Attachment 21367Attachment 21369Attachment 21370

but without success.

Here's the catch: I can't use l'Hopital's Rule. (Headbang)

I would really appreciate some solutions. Thx in advance.

P.S. You don't have to write the whole solutions, just the general idea.
• Apr 5th 2011, 04:46 AM
Sambit
For the "sin" problem, multiply numerator and denominator by $\sqrt{y}+1$ and note that $\lim_{y\to0}\frac{\sin y}{y}=1$, the answer will be $-2$

For the "ln" problem, note that $x$ approaches more rapidly to $0$ than $\ln x$, so the answer becomes $0$.

The other one--algebraic manipulation.
• Apr 5th 2011, 04:50 AM
Soroban

Quote:

$\displaystyle \lim_{x\to4} \frac{x-4}{\sqrt[3]{x+23} - 3}$

First, recall the identity: . $a^3 - b^3 \:=\:(a-b)(a^2+ab + b^2)$

We have: . $\displaystyle f(x) \;=\;\frac{x-4}{(x+23)^{\frac{1}{3}} - 3}$

Multiply numerator and denominator by: $(x+23)^{\frac{2}{3}} + 3(x+23)^{\frac{1}{3}} + 9$

$\displaystyle f(x) \;=\;\frac{x-4}{(x+23)^{\frac{1}{3}} - 3}\cdot\frac{(x+23)^{\frac{2}{3}} + 3(x+23)^{\frac{1}{3}} + 9} {(x+23)^{\frac{2}{3}} + 3(x+23)^{\frac{1}{3}} + 9}$

. . . . $\displaystyle =\;\frac{(x-4)\left[(x+23)^{\frac{2}{3}} + 3(x+23)^{\frac{1}{3}} + 9\right]}{(x+23) - 3^3}$

. . . . $\displaystyle =\;\frac{(x-4)\left[(x+23)^{\frac{2}{3}} + 3(x+23)^{\frac{1}{3}} + 9\right]}{x-4}$

. . . . $\displaystyle =\;(x+23)^{\frac{2}{3}} + 3(x+23)^{\frac{1}{3}} + 9$

Therefore: . $\displaystyle \lim_{x\to4}\left[(x+23)^{\frac{2}{3}} + 3(x+23)^{\frac{1}{3}} + 9\right]$

. . . . . . . . . . $=\; 27^{\frac{2}{3}} + 3\cdot27^{\frac{1}{3}} + 9$

. . . . . . . . . . $=\;9 + 3\cdot3 + 9$

. . . . . . . . . . $= \;27$

• Apr 5th 2011, 05:48 AM
chisigma
Setting $x=1-\xi$ the limit becomes...

$\displaystyle \lim_{x \rightarrow 0} x\ \ln x = \lim_{\xi \rightarrow 1} (1-\xi)\ \ln (1-\xi)$ (1)

Now is...

$\displaystyle \ln (1-\xi) = -\xi - \frac{\xi^{2}}{2} - \frac{\xi^{3}}{3} - ...$ (2)

... so that...

$\displaystyle (1-\xi)\ \ln (1-\xi) = -\xi\ (1-\xi) - \frac{\xi^{2}}{2}\ (1-\xi) - \frac{\xi^{3}}{3}\ (1-\xi) - ...$ (3)

If $\xi = 1$ each term of the 'infinite sum' (3) is 0 so that the limit of 'infinite sum' is 0...

Kind regards

$\chi$ $\sigma$
• Apr 5th 2011, 11:47 PM