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Topics
Pre-Algebra
Mean
Mode
Greatest Common Factor
Least Common Multiple
Order of Operations
Fractions
Mixed Fractions
Prime Factorization
Exponents
Radicals
Algebra
Combine Like Terms
Solve for a Variable
Factor
Expand
Evaluate Fractions
Linear Equations
Quadratic Equations
Inequalities
Systems of Equations
Matrices
Trigonometry
Simplify
Evaluate
Graphs
Solve Equations
Calculus
Derivatives
Integrals
Limits
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Limits
5 problems similar to:
\lim_{ x \rightarrow 0 } 5x
$x→0lim 5x$
Similar Problems from Web Search
Prove that for any c \neq 0 \lim_{x \rightarrow c}{h(x)} does not exist and that \lim_{x \rightarrow 0}{h(x)} does exist.
Prove that for any
$c=0$
$lim_{x→c}h(x)$
does not exist and that
$lim_{x→0}h(x)$
does exist.
https://math.stackexchange.com/questions/334631/prove-that-for-any-c-neq-0-lim-x-rightarrow-chx-does-not-exist-and
Hint: take one sequence that contains only rationals and another one that contains only irrationals (both tending to c\ne 0). For the case of c=0, you can use e.g. that h is continuous at 0 ...
Hint: take one sequence that contains only rationals and another one that contains only irrationals (both tending to
$c=0$
). For the case of
$c=0$
, you can use e.g. that
$h$
is continuous at
$0$
...
Proofs regarding Continuous functions 1
Proofs regarding Continuous functions 1
https://math.stackexchange.com/questions/526691/proofs-regarding-continuous-functions-1
The proof of part a) needs to be modified a bit. You have used the logic that if N \leq f(x) \leq M then xN \leq xf(x) \leq xM. This holds only when x \geq 0. It is better to change the argument ...
The proof of part a) needs to be modified a bit. You have used the logic that if
$N≤f(x)≤M$
then
$xN≤xf(x)≤xM$
. This holds only when
$x≥0$
. It is better to change the argument ...
Use L'Hopital's with this problem?
Use L'Hopital's with this problem?
https://math.stackexchange.com/questions/1419122/use-lhopitals-with-this-problem
Let \displaystyle y=\lim_{x\rightarrow 0^{+}}\left(\frac{1}{x}\right)^{\sin x}\;, Now Let x=0+h\;, Then \displaystyle y=\lim_{h\rightarrow 0}\left(\frac{1}{h}\right)^{\sin h} So \displaystyle \ln(y) = \lim_{h\rightarrow 0}\sin (h)\cdot \ln\left(\frac{1}{h}\right) = -\lim_{h\rightarrow 0}\sin h\cdot \ln(h) = -\lim_{h\rightarrow 0}\frac{\ln(h)}{\csc (h)}\left(\frac{\infty}{\infty}\right) ...
Let
$y=x→0_{+}lim (x1 )_{sinx},$
Now Let
$x=0+h,$
Then
$y=h→0lim (h1 )_{sinh}$
So
$ln(y)=h→0lim sin(h)⋅ln(h1 )=−h→0lim sinh⋅ln(h)=−h→0lim csc(h)ln(h) (∞∞ )$
...
Calculate: \lim_{x \to 0 } = x \cdot \sin(\frac{1}{x})
Calculate:
$lim_{x→0}=x⋅sin(x1 )$
https://math.stackexchange.com/questions/1066434/calculate-lim-x-to-0-x-cdot-sin-frac1x
Your proof is incorrect, cause you used incorrect transform, but it has already been stated. I'll describe way to solve it. \lim_{x \to 0}\frac{\sin(\frac{1}{x})}{\frac{1}{x}} \neq 1 Hint : ...
Your proof is incorrect, cause you used incorrect transform, but it has already been stated. I'll describe way to solve it.
$lim_{x→0}x1 sin(x1 ) =1$
Hint : ...
Prove that f(x) is bounded. Please check my proof.
Prove that
$f(x)$
is bounded. Please check my proof.
https://math.stackexchange.com/q/1052420
Here is another approach: Let L_0 = \lim_{x \downarrow 0} f(x), L_\infty = \lim_{x \to \infty} f(x). By definition of the limit we have some \delta>0 and N>0 such that if x \in (0, \delta), ...
Here is another approach: Let
$L_{0}=lim_{x↓0}f(x),L_{∞}=lim_{x→∞}f(x)$
. By definition of the limit we have some
$δ>0$
and
$N>0$
such that if
$x∈(0,δ)$
, ...
Complex Function limit by investigating sequences
Complex Function limit by investigating sequences
https://math.stackexchange.com/questions/1915934/complex-function-limit-by-investigating-sequences
If a limit as z \to 0 exists, one should be able to plug in any sequence \{ z_n \} going to zero and get the same limit. Limits of sequences are generally easier to work with. So in this case if ...
If a limit as
$z→0$
exists, one should be able to plug in any sequence
${z_{n}}$
going to zero and get the same limit. Limits of sequences are generally easier to work with. So in this case if ...
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Similar Problems
\lim_{ x \rightarrow 0 } 5
$x→0lim 5$
\lim_{ x \rightarrow 0 } 5x
$x→0lim 5x$
\lim_{ x \rightarrow 0 } \frac{2}{x}
$x→0lim x2 $
\lim_{ x \rightarrow 0 } \frac{1}{x^2}
$x→0lim x_{2}1 $
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