Find the limit Lim x → 0 (1 / xsinx + xsin1 / x)

Find the limit Lim x → 0 (1 / xsinx + xsin1 / x)

As shown in the figure

: LIM (xsin1 / x + 1 / xsinx) x tends to 0

The answer is 1
lim(x→0) [xsin(1/x)+(1/x)sinx]
=LIM (x → 0) xsin (1 / x) + LIM (x → 0) SiNx / x, the previous item is (0 × Bounded function), equal to 0
=0+1
=1

Find Lim ┬ (x → 0) ⁡ x ^ 2 + x-tanx / xsinx

lim(x→0) (x^2+x-tanx)/(xsinx)
=lim(x→0) (x^2+x-tanx)/(x^2)
=lim(x→0) 1+(x-tanx)/(x^2)
=lim(x→0) 1+(1-cosx)/(2x)
=1

lim(x-0)tanx-x/x-sinx=

The solution can be obtained by using L'Hospital law
lim(x→0)(tanx-x) /(x-sinx)
= lim(x→0)[(secx)^2-1] /(1-cosx)
= lim(x→0)(1+cosx) / (cosx)^2
= 2.

How to use the intermediate value theorem of continuous function to prove that the function has two zeros, that is, the corresponding equation has two solutions

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Intermediate value theorem of high numbers If f (x) is continuous on [a, b], a Ask for proof.

Because f (x) is continuous on [a, b], there is a maximum value m and a minimum value n on [a, b]; That is, for all x ∈ [a, b], there is n