+6251 \(\displaystyle{\lim_{x\to \infty}}\sqrt{4x^2-x}-\sqrt{9x^2+1}+x\)
\(\displaystyle{\lim_{x\to \infty}}x\left(\sqrt{4-\frac 1 x}-\sqrt{3+\frac 1 x}+1\right)\)
\(\mbox{In the limit we can ignore the }\dfrac 1 x \mbox{ terms}\)
\(\displaystyle{\lim_{x\to \infty}}x\left(\sqrt{4}-\sqrt{3}+1\right)=\displaystyle{\lim_{x\to \infty}}x\left(3-\sqrt{3}\right)\)
\(3 > \sqrt{3} \mbox{ so } \\ \displaystyle{\lim_{x\to \infty}}x\left(3-\sqrt{3}\right)=+\infty \\ \mbox{and so } \\ \displaystyle{\lim_{x\to \infty}}\sqrt{4x^2-x}-\sqrt{9x^2+1}+x=+\infty\)
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lim(sqrt(4x^2-x)-sqrt(9x^2+1)+x),x->infinity
lim_(x->infinity) (sqrt(4 x^2-x)-sqrt(9 x^2+1)+x) = -1/4
Find the following limit:
lim_(x->infinity) (x+sqrt(4 x^2-x)-sqrt(9 x^2+1))
x+sqrt(4 x^2-x)-sqrt(9 x^2+1) = x+sqrt(x (4 x-1))-sqrt(9 x^2+1):
lim_(x->infinity) x+sqrt(x (4 x-1))-sqrt(9 x^2+1)
The limit of x+sqrt(x (4 x-1))-sqrt(9 x^2+1) as x approaches infinity is -1/4:
-1/4
-1/4 = -1/4:
Answer: |
| -1/4
