#How do I solve this limit?

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twilit current
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tawdry mauve
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plug in infinity and you should get zero

solar hull
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If so then you can write sqrt(x^2+1) as x* sqrt(1+1/x^2)

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When x approaches infinity 1/x^2 approaches 0

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So you can do a second order Taylor expansion

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Ie (1+1/x^2)^(1/2)=1+1/2x^2 +o(1/x^2)

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Also x^2/(x+1)=x/(1+1/x)

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=x(1-1/x+1/x^2+o(1/x^2))

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So finally x^2/(1+x)-sqrt(x^2+1)=x-1+1/x-x-1/2x +o(1/x)

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=1/2x -1 +o(1/x)

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1/2x approaches 0 as x approaches infinity so the limit is -1

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Hmmm honestly a second order isn’t even necessary I’m dumb

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(1+1/x^2)^(1/2)=1+o(1/x)

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And 1/(1+1/x)=1-1/x+o(1/x)

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So x/(1+1/x)=x-1+o(1)

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And x*sqrt(1+1/x^2)=x+o(1)

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Thus the expression =-1+o(1)=>f(x)~-1 for x approaching infinity with f the function which your trying to find the limit

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Ie it’s limit is -1

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Or you could say x^2=(x^2-1)+1

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Thus x^2/(1+x)=(x^2-1)/(x+1)+1/(x+1)

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=(x+1)(x-1)/(x+1) -0 when x approaches infinity

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=x-1

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Thus x^2/(x+1)-sqrt(x^2+1)=x-1-sqrt(x^2+1)~x-1-sqrt(x^2)~-1

twilit current
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😮 Sorry for the late reply. Yes I know the taylor series with limits. Thank you so much ❤️

unkempt wagon
tender vapor
twilit current
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Is there another way without taylor?

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@solar hull

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Tried doing it where I multiply numerator and denom with a root

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and then divide by x

solar hull
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Actually you can calculate the limit using derivative rules

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Like this

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Using the definition of the derivative

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This doesn’t use Taylor expansion

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@twilit current

twilit current
# solar hull

oh wow! Thank you!! 🙂 Your help is very much appreciated

small shoreBOT
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