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education/math/MATH1210 (calc 1)/Limits.md

@@ -88,7 +88,7 @@ The above problem can be solved more easily *without* L'Hospital's rule, the lea
 L'Hospital's rule **cannot** be used in any other circumstance. 
 
 ## Examples
-1. $\lim_{x \ to 0} \dfrac{7^x - 5^x}{2x}$
+1. $\lim_{x \to 0} \dfrac{7^x - 5^x}{2x}$
 2. $= \lim_{x \ to 0}\dfrac{7^x \ln(7) -5^x(\ln(5)}{2}$
 3. $= \dfrac{\ln(7) - \ln(5)}{2}$ 
 # Indeterminate form $(0 * \infty)$
@@ -96,4 +96,4 @@ If the $\lim_{x \to a}f(x) = 0$  and $\lim_{x\to a} g(x) = \infty$ then $\lim_{x
 
 To evaluate an indeterminate product ($0 * \infty$), use algebra to convert the product to an equivalent quotient and then use L'Hopsital's Rule.
 
-$$ \lim_{x \to 0^+} x\ln(x) = \lim_{x \to 0^+}\dfrac{\ln x}{\dfrac{1}{x}} = \\lim_{x \to 0^+} \dfrac{1/x}{-1/(x^2)} = \lim_{x \to 0^+} -x = 0 $$
+$$ \lim_{x \to 0^+} x\ln(x) = \lim_{x \to 0^+}\dfrac{\ln x}{\dfrac{1}{x}} = \lim_{x \to 0^+} \dfrac{1/x}{-1/(x^2)} = \lim_{x \to 0^+} -x = 0 $$